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ELEMENTARY  SYNTHETIC  GEOMETRY. 


ELEMENTARY 

SYNTHETIC    GEOMETRY 


OF  THE 


POINT,   LINE   AND   CIRCLE   IN 
THE   PLANE 


BY 

N.   F.   DUPUIS,   M.A,   F.R.S.C. 

PROFESSOR   OF    PURE   MATHEMATICS   IN    THE   UNIVERSITY  OF   QUEEN's 


COLLEGE,    KINGSTON,    CANADA. 


MACMILLAN    AND    CO. 

AND   NEW   YORK. 

1889. 
[Aii  rights  reset  fed] 


CAJORl 


PREFACE. 

The  present  work  is  a  result  of  the  Author's  experience 
in  teaching  Geometry  to  Junior  Classes  in  the  University 
for  a  series  of  years.  It  is  not  an  edition  of  "  Euclid's 
Elements,"  and  has  in  fact  little  relation  to  that  cele- 
brated ancient  work  except  in  the  subject  matter. 

The  work  differs  also  from  the  majority  of  modern 
treatises  on  Geometry  in  several  respects. 

The  point,  the  line,  and  the  curve  lying  in  a  common 
plane  are  taken  as  the  geometric  elements  of  Plane 
Geometry,  and  any  one  of  these  or  any  combination  of 
them  is  defined  as  a  geometric  plane  figure.  Thus  a 
triangle  is  not  the  three-cornered  portion  of  the  plane 
inclosed  within  its  sides,  but  the  combination  of  the 
three  points  and  three  lines  forming  what  are  usually 
termed  its  vertices  and  its  sides  and  sides  produced. 

This   mode   of   considering    geometric    figures    leads 

911298 


VI  PREFACE. 

naturally  to  the  idea  of  a  figure  as  a  locus,  and  con- 
sequently prepares  the  way  for  the  study  of  Cartesian 
Geometry.  It  requires,  however,  that  a  careful  distinction 
be  drawn  between  figures  which  are  capable  of  super- 
position and  those  which  are  equal  merely  in  area. 
The  properties  of  congruence  and  equality  are  accord- 
ingly carefully  distinguished. 

The  principle  of  motion  in  the  transformation  of 
geometric  figures,  as  recommended  by  Dr.  Sylvester, 
and  as  a  consequence  the  principle  of  continuity  are 
freely  employed,  and  an  attempt  is  made  to  generalize 
all  theorems  which  admit  of  generalization. 

An  endeavour  is  made  to  connect  Geometry  with 
Algebraic  forms  and  symbols.,  (i)  by  an  elementary 
study  of  the  modes  of  representing  geometric  ideas  in 
the  symbols  of  Algebra,  and  (2)  by  determining  the 
consequent  geometric  interpretation  which  is  to  be  given 
to  each  interpretable  algebraic  form.  The  use  of  such 
forms  and  symbols  not  only  shortens  the  statements 
of  geometric  relations  but  also  conduces  to  greater 
generality. 

In  dealing  with  proportion  the  method  of  measures 
is  employed  in  preference  to  that  of  multiples  as  being 


PREFACE.  Vll 

equally  accurate,  easier  of  comprehension,  and  more 
in  line  with  elementary  mathematical  study.  In  dealing 
with  ratio  I  have  ventured,  when  comparing  two  finite 
lines,  to  introduce  Hamilton's  word  tensor  as  seeming  to 
me  to  express  most  clearly  what  is  meant. 

After  treating  of  proportion  I  have  not  hesitated  to 
employ  those  special  ratios  known  as  trigonometric  func- 
tions in  deducing  geometric  relations. 

In  the  earlier  parts  of  the  work  Constructive  Geometry 
is  separated  from  Descriptive  Geometry,  and  short 
descriptions  are  given  of  the  more  important  geometric 
drawing-instruments,  having  special  reference  to  the 
geometric  principles  of  their  actions. 

Parts  IV.  and  V.  contain  a  synthetic  treatment  of  the 
theories  of  the  mean  centre,  of  inverse  figures,  of  pole 
and  polar,  of  harmonic  division,  etc.,  as  applied  to  the 
line  and  circle ;  and  it  is  believed  that  a  student  who 
becomes  acquainted  with  these  geometric  extensions  in 
this  their  simpler  form  will  be  greatly  assisted  in  the 
wider  discussion  of  them  in  analytical  conies.  Through- 
out the  whole  work  modern  terminology  and  modern 
processes  have  been  used  with  the  greatest  freedom, 
regard  being  had  in  all  cases  to  perspicuity. 


viii  PREFACE. 

As  is  evident  from  what  has  been  said,  the  whole 
intention  in  preparing  the  work  has  been  to  furnish  the 
student  with  that  kind  of  geometric  knowledge  which 
may  enable  him  to  take  up  most  successfully  the  modern 
works  on  Analytical  Geometry. 

N.  F.  D 

Queen's  Coli.k;ge, 

Kingston,  Canada. 


CONTENTS. 

PART   I. 

I'AGE 

Section  L— The  Line  and  Point.  Section  II.— Two 
Lines — Angles.  Section  III. — Three  or  more 
Lines  and  Determined  Points — The  Triangle. 
Section  IV.— Parallels.  Section  V.— The  Circle. 
Section  VI. — Constructive  Geometry,  .        .        .        i 

PART   II. 

Section  I.-— Comparison  of  Areas.  Section  II.-- 
iVIeasurement  of  Lengths  and  Areas.  Section 
111. — Geometric  Interpretation  of  Algebraic  Forms. 
Section  IV. — Areal  Relations — Squares  and  Rect- 
angles.    Section  V. — Constructive  Geometry,      .      91 

PART  III. 

Section  I.  —  Proportion  amongst  Line  -  Segments. 
Section  II.  —  Functions  of  Angles  and  their 
Applications  in  Geometry, 147 


X  CONTENTS. 

PART    IV. 

PAGE 

Section  I. — Geometric  Extensions.  Section  II. — 
Centre  of  Mean  Position.  Section  III.— Col- 
linearity  and  Concurrence.  Section  IV.— Inver- 
sion and  Inverse  Figures.  Section  V. — Pole  and 
Polar.  Section  VI.— The  Radical  Axis.  Sec- 
tion VII. — Centres  and  Axes  of  Perspective  or 
Similitude, 178 

PART  V. 

Section  I.— Anharmonic  Division.  Section  II. — 
Harmonic  Ratio.  Section  III. — Anharmonic 
Properties.  Section  IV.— Polar  Reciprocals  and 
Reciprocation.  Section  V. — Homography  and 
Involution, 252 


ELEMENTARY   SYNTHETIC    GEOMETRY. 


PART    I 


GENERAL   CONSIDERATIONS. 

1°.  A  statement  which  explains  the  sense  in  which  some 
word  or  phrase  is  employed  is  2i  definition. 

A  definition  may  select  some  one  meaning  out  of  several 
attached  to  a  common  word,  or  it  may  introduce  some 
technical  term  to  be  used  in  a  particular  sense. 

Some  terms,  such  as  space,  straight,  direction,  etc.,  which 
express  elementary  ideas  cannot  be  defined. 

2°.  Def. — A  Theorem  is  the  formal  statement  of  some 
mathematical  relation. 

A  theorem  may  be  stated  for  the  purpose  of  being  sub- 
sequently proved,  or  it  may  be  deduced  from  some  previous 
course  of  reasoning. 

In  the  former  case  it  is  called  a  Proposition^  that  is,  some- 
thing proposed,  and  consists  of  {a)  the  statement  or  enuncia- 
tion of  the  theorem,  and  {b)  the  argument  or  proof.  The 
purpose  of  the  argument  is  to  show  that  the  truth  of  the 
theorem  depends  upon  that  of  some  preceding  theorem 
whose  truth  has  already  been  established  or  admitted. 

Ex.  "  The  sum  of  two  odd  numbers  is  an  even  number  " 
is  a  theorem. 

3°.  A  theorem  so  elementary  as  to  be  generally  accepted  as 
true  without  any  formal  proof,  is  an  axiom. 
«  A 


2  SYNTHETIC   GEOMETRY. 

Mathematical  axioms  are  general  or  particular,  that  is, 
they  apply  to  the  whole  science  of  mathematics,  or  have 
special  applications  to  some  department. 
The  principal  general  axioms  are  : — 

i.  The  whole  is  equal  to  the  sum  of  all  its  parts,  and 

therefore  greater  than  any  one  of  its  parts. 
ii.  Things  equal  to  the  same  thing  are  equal  to  one 

another, 
iii.  If  equals  be  added  to  equals  the  sums  are  equal, 
iv.  If  equals  be  taken  from  equals  the  remainders  are 

equal. 
V.  If  equals  be  added  to  or  taken  from  unequals  the 

results  are  unequal, 
vi.  If  unequals  be  taken  from  equals  the  remainders  are 

unequal, 
vii.  Equal  multiples  of  equals  are  equal ;  so  also  equal 
submultiples  of  equals  are  equal. 
The  axioms  which  belong  particularly  to  geometry  will 
occur  in  the  sequel. 

4°.  The  statement  of  any  theorem  may  be  put  into  the 
hypothetical  form,  of  which  the  type  is— 
If  A  is  B  then  C  is  D. 
The  first  part  "  if  A  is  B  "  is  called  the  hypothesis^  and  the 
second  part  "  then  C  is  D  "  is  the  conclusion. 

Ex.  The  theorem  "  The  product  of  two  odd  numbers  is 
an  odd  number  "  can  be  arranged  thus  : — 

Hyp.     If  two  numbers  are  each  an  odd  number. 
Concl.  Then  their  product  is  an  odd  number. 

5°.  The  statement  "  If  A  is  B  then  C  is  D  "  may  be  im- 
mediately put  into  the  form — 

If  C  is  not  D  then  A  is  not  B, 
which  is  called  the  contrapositive  of  the  former. 

The  truth  of  a  theorem  establishes  the  truth  of  its  contra- 


GENERAL    CONSIDERATIONS.  3 

positive,  and  vice  versa,  and  hence  if  either  is  proved  the 
other  is  proved  also. 

6°.  Two  theorems  are  converse  to  one  another  when  the 
hypothesis  and  conclusion  of  the  one  are  respectively  the 
conclusion  and  hypothesis  of  the  other. 

Ex.  If  an  animal  is  a  horse  it  has  four  legs. 

Converse.  If  an  animal  has  four  legs  it  is  a  horse. 

As  is  readily  seen  from  the  foregoing  example,  the  truth  of 
a  theorem  does  not  necessarily  establish  the  truth  of  its  con- 
verse, and  hence  a  theorem  and  its  converse  have  in  general 
to  be  proved  separately.  But  on  account  of  the  peculiar 
relation  existing  between  the  two,  a  relation  exists  also 
between  the  modes  of  proof  for  the  two.  These  are  known 
as  the  direct  and  indirect  modes  of  proof.  And  if  any 
theorem  which  admits  of  a  converse  can  be  proved  directly 
its  converse  can  usually  be  proved  indirectly.  Examples 
will  occur  hereafter. 

7°.  Many  geometric  theorems  are  so  connected  with  their 
converses  that  the  truth  of  the  theorems  establishes  that  of 
the  converses,  and  vice  versa. 

The  necessary  connection  is  expressed  in  the  Rule  of 
Identity i  its  statement  being  : — 

If  there  is  but  07ie  X  and  one  V,  and  if  it  is  proved 
that  X  is  V,  then  it  follows  that  V  is  X. 

Where  X  and  Y  stand  for  phrases  such  as  may  form  the 
hypotheses  or  conclusions  of  theorems,  and  the  "  is  "  between 
them  is  to  be  variously  interpreted  as  "  equal  to,"  "  corre- 
sponds to,"  etc. 

Ex.  Of  two  sides  of  a  triangle  only  one  can  be  the  greater, 
and  of  the  two  angles  opposite  these  sides  only  one  can  be 
the  greater.  Then,  if  it  is  proved  that  the  greater  side  is 
opposite  the  greater  angle  it  follows  that  the  greater  angle  is 
opposite  the  greater  side. 


4  SYNTHETIC  GEOMETRY. 

In  this  example  there  is  but  one  X  (the  greater  side)  and 
one  Y  (the  greater  angle),  and  as  X  is  (corresponds  to  or  is 
opposite)  Y,  therefore  Y  is  (corresponds  to  or  is  opposite)  X. 

8°.  A  Corollary  is  a  theorem  deduced  from  some  other 
theorem,  usually  by  some  qualification  or  restriction,  and 
occasionally  by  some  amplification  of  the  hypothesis.  Or  a 
corollary  may  be  derived  directly  from  an  axiom  or  from  a 
definition. 

As  a  matter  of  course  no  sharp  distinction  can  be  drawn 
between  theorems  and  corollaries. 

Ex.  From  the  theorem,  "  The  product  of  two  odd  numbers 
is  an  odd  number,"  by  making  the  two  numbers  equal  we 
obtain  as  a  corollary,  "  The  square  of  an  odd  number  is  an 
odd  number." 

Exercises. 

State  the  contrapositives  and  the  converses  of  the  follow- 
ing theorems  : — 

1.  The  sum  of  two  odd  numbers  is  an  even  number. 

2.  A  diameter  is  the  longest  chord  in  a  circle. 

3.  Parallel  lines  never  meet. 

4.  Every  point  equidistant  from  the  end-points  of  a  line- 

segment  is  on  the  right  bisector  of  that  segment. 


SECTION   I. 

THE   LINE   AND   POINT. 

9°.  Space  may  be  defined  to  be  that  which  admits  of 
length  or  distance  in  every  direction ;  so  that  length  and 
direction  are  fundamental  ideas  in  studying  the  geometric 
properties  of  space. 


THE   LINE  AND   POINT.  5 

Every  material  object  exists  in,  and  is  surrounded  by 
space.  The  limit  which  separates  a  material  object  from 
the  space  which  surrounds  it,  or  which  separates  the  space 
occupied  by  the  object  from  the  space  not  occupied  by  it,  is 
a  surface. 

The  surface  of  a  black-board  is  the  limit  which  separates 
the  black-board  from  the  space  lying  without  it.  This  surface 
can  have  no  thickness,  as  in  such  a  case  it  would  include  a 
part  of  the  board  or  of  the  space  without  or  of  both,  and 
would  not  be  the  dividing  Hmit. 

io°.  A  flat  surface,  as  that  of  a  black-board,  is  a  plane 
surface,  or  a  Plaiie, 

Pictures  of  geometric  relations  drawn  on  a  plane  surface 
as  that  of  a  black-board  are  usually  called  Plane  Geometric 
Figures^  because  these  figures  lie  in  or  on  a  plane. 

Some  such  figures  are  known  to  every  person  under  such 
names  as  "  triangle,"  "  square,"  "  circle,"  etc. 

11°.  That  part  of  mathematics  which  treats  of  the  properties 
and  relations  of  plane  geometric  figures  is  Plane  Geometry. 
Such  is  the  subject  of  this  work. 

The  plane  upon  which  the  figures  are  supposed  to  lie  will 
be  referred  to  as  the  plane,  and  unless  otherwise  stated  all 
figures  will  be  supposed  to  lie  in  or  on  the  same  plane. 

12°.  The  Line.  When  the  crayon  is  drawn  along  the 
black-board  it  leaves  a  visible  mark.  This  mark  has  breadth 
and  occupies  some  of  the  surface  upon  which  it  is  drawn, 
and  by  way  of  distinction  is  called  a  physical  line.  By 
continually  diminishing  the  breadth  of  the  physical  line  we 
make  it  approximate  to  the  geometric  line.  Hence  we  may 
consider  the  geometric  line  as  being  the  limit  towards  which 
a  physical  line  approaches  as  its  breadth  is  continually 
diminished.  We  may  consequently  consider  a  geometric 
line  as  length  abstracted  from  every  other  consideration. 


6  SYNTHETIC   GEOMETRY. 

This  theoretic  relation  of  a  geometric  line  to  a  physical 
one  is  of  some  importance,  as  whatever  is  true  for  the 
physical  line,  independently  of  its  breadth,  is  true  for  the 
geometric  line.  And  hence  arguments  in  regard  to  geometric 
lines  may  be  replaced  by  arguments  in  regard  to  physical 
lines,  if  from  such  arguments  we  exclude  everything  that 
would  involve  the  idea  of  breadth. 

The  diagrams  employed  to  direct  and  assist  us  in  geo- 
metric investigations  are  formed  of  physical  lines,  but  they 
may  equally  well  be  supposed  to  be  formed  of  threads,  wires 
or  light  rods,  if  we  do  not  involve  in  our  arguments  any  idea 
of  the  breadth  or  thickness  of  the  lines,  threads,  wires  or 
rods  employed. 

In  the  practical  applications  of  Geometry  the  diagrams 
frequently  become  material  or  represent  material  objects. 
Thus  in  Mechanics  we  consider  such  things  as  levers,  wedges, 
wheels,  cords,  etc.,  and  our  diagrams  become  representations 
of  these  things. 

A  pulley  or  wheel  becomes  a  circle,  its  arms  become  radii 
of  the  circle,  and  its  centre  the  centre  of  the  circle  ;  stretched 
cords  become  straight  lines,  etc. 

13°.  The  Point.  A  point  marks  position,  but  has  no  size. 
The  intersection  of  one  line  by  another  gives  a  point,  called 
the  point  of  intersection. 

If  the  lines  are  physical,  the  point  is  physical  and  has  some 
size,  but  when  the  lines  are  geometric  the  point  is  also 
geometric. 

14°.  Straight  Line.  For  want  of  a  better  definition  we 
may  say  that  a  straight  line  is  one  of  which  every  part  has 
the  same  direction.  For  every  part  of  a  line  must  have  some 
direction,  and  when  this  direction  is  common  to  all  the  parts 
of  the  line,  the  line  is  straight. 

The  word  "  direction  "  is  not  in  itself  definable,  and  when 
applied  to  a  line  in  the  absolute  it  is  not  intelligible.     But 


THE   LINE   AND    POINT.  7 

every  person  knows  what  is  meant  by  such  expressions  as 
"the  same  direction,"  "opposite  direction,"  etc.,  for  these 
express  relations  between  directions,  and  such  relations  are 
as  readily  comprehended  as  relations  between  lengths  or 
other  magnitudes. 

The  most  prominent  property,  and  in  fact  the  distinctive 
property  of  a  straight  line,  is  the  absolute  sameness  which 
characterizes  all  its  parts,  so  that  two  portions  of  the  same 
straight  line  can  differ  from  one  another  in  no  respect  except 
in  length. 

Def. — A  plane  figure  made  up  of  straight  lines  only  is 
called  a  rectilinear  figure. 

15°.  A  Curve  is  a  line  of  which  no  part  is  straight ;  or  a 
curve  is  a  line  of  which  no  two  adjacent  parts  have  the  same 
direction. 

The  most  common  example  of  a  curve  is  a  circle  or  portion 
of  a  circle. 

Henceforward,  the  word  "line,"  unless  otherwise  qualified, 
will  mean  a  straight  line. 

16°.  The  "rule"  or  "straight-edge"  is  a  strip  of  wood, 
metal,  or  other  solid  with  one  ^d^g^^  made  straight.  Its 
common  use  is  to  guide  the  pen  or  pencil  in  drawing  lines 
in  Practical  Geometry. 

17°.  A  Plane  is  a  surface  such  that  the  line  joining  any  two 
arbitrary  points  in  it  coincides  wholly  with  the  surface. 

The  planarity  of  a  surface  may  be  tested  by  applying  the 
rule  to  it.  If  the  rule  touches  the  surface  at  some  points  and 
not  at  others  the  sujface  is  not  a  plane.  But  if  the  rule 
touches  the  surface  throughout  its  whole  length,  and  in  every 
position  and  direction  in  which  it  can  be  applied,  the  surface 
is  a  plane. 

The  most  accurately  plane  artificial  surface  known  is 
probably   that   of  a   well-formed   plane  mirror.     Examina- 


8  SYNTHETIC   GEOMETRY. 

tion  of  the  images  of  objects  as  seen  in  such  mirrors  is 
capable  of  detecting  variations  from  the  plane,  so  minute  as 
to  escape  all  other  tests. 

1 8°.  A  surface  which  is  not  plane,  and  which  is  not  com- 
posed of  planes,  is  a  curved  surface.  Such  is  the  surface  of 
a  sphere,  or  cylinder. 

19°.  The  point,  the  line,  the  curve,  the  plane  and  the 
curved  surface  are  the  elements  which  go  to  make  up  geo- 
metric figures. 

Where  a  single  plane  is  the  only  surface  concerned,  the 
point  and  line  lie  in  it  and  form  a  plane  figure.  But  where 
more  than  one  plane  is  concerned,  or  where  a  curved  surface 
is  concerned,  the  figure  occupies  space,  as  a  cube  or  a  sphere, 
and  is  called  a  spatial  figure  or  a  solid. 

The  study  of  spatial  figures  constitutes  Solid  Geometry,  or 
the  Geometry  of  Space,  as  distinguished  from  Plane  Geometry. 

20°.  Given  Point  and  Line.  A  point  or  line  is  said  to  be 
given  when  we  are  made  to  know  enough  about  it  to  enable 
us  to  distinguish  it  from  every  other  point  or  line ;  and  the 
data  which  give  a  point  or  line  are  commonly  said  to 
determine  it. 

A  similar  nomenclature  applies  to  other  geometric  ele- 
ments. 

The  statement  that  a  point  or  line  lies  in  a  plane  does  not 
give  it,  but  a  point  or  line  placed  in  the  plane  for  future 
reference  is  considered  as  being  given.  Such  a  point  is 
usually  called  an  origin^  and  such  a  line  a  datum  line^  an 
initial  lijie,  a  pri^ne  vector ^  etc. 

21°.  Def.  I. — A  line  considered  merely  as  a  geometric 
element,  and  without  any  limitations,  is  an  indefinite  line. 

2. — A  limited  portion  of  a  line,  especially  when  any  refer- 
ence is  had  to  its  length,  is  2i  finite  line,  or  a  line-segment,  or 
simply  a  segment. 


THE   LINE  AND   POINT.  9 

That  absolute  sameness  (14°)  which   characterizes  every 
part  of  a  line  leads  directly  to  the  following  conclusions  : — 
(i)  No  distinction  can  be  made  between  any  two  segments 
of  the  same  line  equal  in  length,  except  that  of  position 
in  the  line. 

(2)  A  line  cannot  return  into,  or  cross  itself. 

(3)  A  line  is  not  necessarily  limited  in  length,  and  hence, 

in  imagination,  we  may  follow  a  line  as  far  as  we 
please  without  coming  to  any  necessary  termination. 

This  property  is  conveniently  expressed  by  saying 
that  a  line  extends  to  infinity. 
3. — The  hypothetical  end-points  of  any  indefinite  line  are 
said  to  be  points  at  infinity.     All  other  points  are  finite 
points. 

22°.  Notation.  A  point  is  denoted  by  a  single  letter  where- 
ever  practicable,  as  "  the  point  A." 

An  indefinite  line  is  also  denoted  by  a  single  letter  as  "the 
line  L,"  but  in  this  case  the  letter  l 

has  no  reference  to  any  point.  a  b 

A  segment  is  denoted  by  naming  its  end  points,  as  the 
"  segment  AB,"  where  A  and  B  are  the  e7id  points.  This  is 
a  biliteral.,  or  two-letter  notation. 

A  segment  is  also  denoted  by  a  single  letter,  when  the 
limits  of  its  length  are  supposed  to  be  known,  as  the  "  seg- 
ment rt."     This  is  a  uniliteral,  or  one-letter  notation. 

The  term  "segment"  involves  the  notion  of  some  finite 
length.  When  length  is  not  under  consideration,  the  term 
"  line  "  is  preferred. 

Thus  the  "  Hne  AB  "  is  the  indefinite  line  having  A  and  B 
as  two  points  upon  it.  But  the  "segment  AB"  is  that  portion 
of  the  line  which  lies  between  A  and  B. 

23°.  In  dealing  with  a  hne-segment,  we  frequently  have  to 
consider  other  portions  of  the  indefinite  line  of  which  the 
segment  is  a  part. 


lO  SYNTHETIC  GEOMETRY. 

As  an  example,  let  it  be  required  to  divide  the  segment  AB 

, __, , into  two  parts  whereof  one  shall 

^       ^    ^  c  be  twice  as  long  as  the  other. 

To  do  this  we  put  C  in  such  a  position  that  it  may  be  twice  as 
far  from  one  of  the  end-points  of  the  segment,  A  say,  as  it  is 
from  the  other,  B.  But  on  the  indefinite  line  through  A  and 
B  we  may  place  C  so  as  to  be  twice  as  far  from  A  as  from  B. 
So  that  we  have  two  points,  C  and  C,  both  satisfying  the 
condition  of  being  twice  as  far  from  A  as  from  B. 

Evidently,  the  point  C  does  not  divide  the  segment  AB  in 
the  sense  commonly  attached  to  the  word  divide.  But  on 
account  of  the  similar  relations  held  by  C  and  C  to  the  end- 
points  of  the  segment,  it  is  convenient  and  advantageous  to 
consider  both  points  as  dividing  the  segment  AB. 

When  thus  considered,  C  is  said  to  divide  the  segment 
internally  and  C  to  divide  it  externally  in  the  same  manner. 

24°.  Axiom. — Through  a  given  point  only  one  line  can 
pass  in  a  given  direction. 

Let  A  be  the  given  point,  and  let  the  segment  AP  mark 

— V.    the  given  direction.     Then,  of  all  the  lines 

that  can  pass  through  the  point  A,  only  one 
can  have  the  direction  AP,  and  this  one  must  lie  along  and 
coincide  with  AP  so  as  to  form  with  it  virtually  but  one  line. 

Cor.  I.  A  finite  point  and  a  direction  determine  one 
line. 

Cor.  2.  Two  given  finite  points  determine  one  line.  For,  if 
A  and  P  be  the  points,  the  direction  AP  is  given,  and  hence 
the  line  through  A  and  having  the  direction  AP  is  given. 

Cor.  3.  Two  lines  by  their  intersection  determine  one  finite 
point.  For,  if  they  determined  two,  they  would  each  pass 
through  the  same  two  points,  which,  from  Cor.  2,  is  impossible. 

Cor.  4.  Another  statement  of  Cor.  2  is — Two  lines  which 
have  two  points  in  common  coincide  and  form  virtually  but 
one  line. 


THE   LINE   AND    POINT.  II 

25°.  Axiom. — A  straight  line  is  the  shortest  distance  be- 
tween two  given  points. 

Although  it  is  possible  to  give  a  reasonable  proof  of  this 
axiom,  no  amount  of  proof  could  make  its  truth  more 
apparent. 

The  following  will  illustrate  the  axiom.  Assume  any  two 
points  on  a  thread  taken  as  a  physical  line.  By  separating 
these  as  far  as  possible,  the  thread  takes  the  form  which  we 
call  straight,  or  tends  to  take  that  form.  Therefore  a  straight 
finite  line  has  its  end-points  further  apart  than  a  curved  line 
of  equal  length.  Or,  a  less  length  of  line  will  reach  from  one 
given  point  to  another  when  the  line  is  straight  than  when  it 
is  curved. 

Def. — The  distance  between  two  points  is  the  length  of  the 
segment  which  connects  them  or  has  them  as  end-points. 

26°.  Superpositiojt. — Comparison  of  Figjcres.—V\[e  assume 
that  space  is  homogeneous,  or  that  all  its  parts  are  alike,  so 
that  the  properties  of  a  geometric  figure  are  independent  of  its 
position  in  space.  And  hence  we  assume  that  a  figure  may 
be  supposed  to  be  moved  from  place  to  place,  and  to  be  turned 
around  or  over  in  any  way  without  undergoing  any  change 
whatever  in  its  form  or  properties,  or  in  the  relations  existing 
between  its  several  parts. 

The  imaginary  placing  of  one  figure  upon  another  so 
as  to  compare  the  two  is  called  siiperpositioti.  By  superposi- 
tion we  are  enabled  to  compare  figures  as  to  their  equality  or 
inequality.  If  one  figure  can  be  superimposed  upon  another 
so  as  to  coincide  with  the  latter  in  every  part,  the  two  figures 
are  necessarily  and  identically  equal,  and  become  virtually 
one  figure  by  the  superposition. 

27°.  Two  line-segments  can  be  compared  with  respect  to 
length  only.  Hence  a  line  is  called  a  magnitude  of  one 
dimension. 

Two  segments  are  equal  when  the  end-points  of  one  can  be 


12  SYNTHETIC   GEOMETRY. 

made  to  coincide  with  the  end-points  of  the  other  by  super- 
position. 

28°.  Def. — The  stun  of  two  segments  is  that  segment  which 
is  equal  to  the  two  when  placed  in  line  with  one  end-point  in 
each  coincident. 

Let  AB  and  DE  be  two  segments,  and  on  the  line  of  which 

D E  AB  is  a  segment  let   BC   be   equal  to 

^ DE.      Then    AC    is    the   stem    of   AB 

"a  b  c  and  DE. 

This  is  expressed  symbolically  by  writing 
AC  =  AB  +  DE, 
where  =  denotes  equality  in  length,  and  +  denotes  the 
placing  of  the  segments  AB  and  DE  in  line  so  as  to  have  one 
common  point  as  an  end-point  for  each.  The  interpretation 
of  the  whole  is,  that  AC  is  equal  in  length  to  AB  and  DE 
together. 

29°.  Def, — The  difference  between  two  segments  is  the 
segment  which  remains  when,  from  the  longer  of  the  segments, 
a  part  is  taken  away  equal  in  length  to  the  shorter. 

Thus,  if  AC  and  DE  be  two  segments  of  which  AC  is  the 
longer,  and  if  BC  is  equal  to  DE,  then  AB  is  the  difference 
between  AC  and  DE. 

This  is  expressed  symbolically  by  writing 
AB=AC-DE, 
which  is  interpreted  as  meaning  that  the   segment  AB  is 
shorter  than  AC  by  the  segment  DE. 

Now  this  is  equivalent  to  saying  that  AC  is  longer  than 
AB  by  the  segment  DE,  or  that  AC  is  equal  to  the  sum  of 
AB  and  DE. 

Hence  when  we  have    AB  =  AC-  DE 
we  can  write  AC  =  AB-l-DE. 

We  thus  see  that  in  using  these  algebraic  symbols,  =,  +, 
and  -,  a  term,  as  DE,  may  be  transferred  from  one  side  of 


THE   LINE  AND   POINT.  1 3 

the  equation  to  another  by  changing  its  sign  from  +  to  -  or 
vice  versa. 

Owing  to  the  readiness  with  which  these  symbolic  expres- 
sions can  be  manipulated,  they  seem  to  represent  simple  alge- 
braic relations,  hence  beginners  are  apt  to  think  that  the  work- 
ing rules  of  algebra  must  apply  to  them  as  a  matter  of 
necessity.  It  must  be  remembered,  however,  that  the  formal 
rules  of  algebra  are  founded  upon  the  properties  of  numbers, 
and  that  we  should  not  assume,  without  examination,  that 
these  rules  apply  without  modification  to  that  which  is  not 
number. 

This  subject  will  be  discussed  in  Part  II. 

30°.  Def. — That  point,  in  a  line-segment,  which  is  equi- 
distant from  the  end-points  is  the  7niddle  point  of  the  segment. 

It  is  also  called  the  internal  point  of  bisection  of  the  seg- 
ment, or,  when  spoken  of  alone,  simply  \h^  point  of  bisection. 

Exercises. 

1.  If  two  segments  be  in  line  and  have  one  common  end- 

point,  by  what  name  will  you  call  the  distance  between 
their  other  end-points  .'* 

2.  Obtain  any  relation  between  "  the  sum  and  the  differ- 

ence" of  two  segments  and  "the  relative  directions  "of 
the  two  segments,  they  being  in  line. 

3.  A  given  line-segment  has  but  one  middle  point 

4.  In  Art.  23°,  if  C  becomes  the  middle  point  of  AB,  what 

becomes  of  C 1 

5.  In  Art.  30°  the  internal  point  of  bisection  is  spoken  of. 

What  meaning  can  you  give  to  the  "  external  point  of 
bisection  "  1 


14  SYNTHETIC   GEOMETRY. 

SECTION    11. 

RELATIONS  OF  TWO  LINES.— ANGLES. 

31°.  When  two  lines  have  not  the  same  direction  they  are 
said  to  make  an  angle  with  one  another,  and  an  angle  is  a 
c   differejice  in  direction. 

'^       Illustratio7i.'—'L^\.  A  and  B 

represent  two  stars,  and  E  the 

A'        *A  position  of  an  observer's  eye. 

Since  the  lines  EA  and  EB,  which  join  the  eye  and  the 

stars,  have  not  the  same  direction  they  make  an  angle  with 

one  another  at  E. 

1.  If  the  stars  appear  to  recede  from  one  another,  the  angle 
at  E  becomes  greater.  Thus,  if  B  moves  into  the  position  of 
C,  the  angle  between  EA  and  EC  is  greater  than  the  angle 
between  EA  and  EB, 

Similarly,  if  the  stars  appear  to  approach  one  another,  the 
angle  at  E  becomes  smaller ;  and  if  the  stars  become  coinci- 
dent, or  situated  in  the  same  line  through  E,  the  angle  at  E 
vanishes. 

Hence  an  angle  is  capable  of  continuous  increase  or 
diminution,  and  is  therefore  a  magnitude.  And,  being 
magnitudes,  angles  are  capable  of  being  compared  with  one 
another  as  to  greatness,  and  hence,  of  being  measured. 

2.  If  B  is  moved  to  B',  any  point  on  EB,  and  A  to  A',  any 
point  on  EA,  the  angle  at  E  is  not  changed.  Hence  increas- 
ing or  diminishing  one  or  both  of  the  segments  which  form 
an  angle  does  not  affect  the  magnitude  of  the  angle. 

Hence,  also,  there  is  no  community  in  kind  between  an 
angle  and  a  line-segment  or  a  line. 

Hence,  also,  an  angle  cannot  be  measured  by  means  of 
line-segments  or  lines. 


RELATIONS  OF   TWO   LINES. — ANGLES.  1 5 

32°.  Def. — A  line  which  changes  its  direction  in  a  plane 
while  passing  through  a  fixed  point  in  the  plane  is  said  to 
rotate  about  the  point. 

The  point  about  which  the  rotation  takes  place  is  the  pole^ 
and  any  segment  of  the  rotating  line,  having  the  pole  as  an 
end-point,  is  a  radius  vector. 

Let  an  inextensible  thread  fixed  at  O 
be  kept  stretched  by  a  pencil  at  P. 
Then,  when  P  moves,  keeping  the 
thread  straight,  OP  becomes  a  radius 
vector  rotating  about  the  pole  O.  ^  "^ 

When  the  vector  rotates  from  direction  OP  to  direction 
OP'  it  describes  the  angle  between  OP  and  OP'.  Hence  we 
have  the  following  : — 

Def.  I.  —  The  angle  between  two  lines  is  the  I'otation  neces- 
sary to  bring  one  of  the  lines  into  the  direction  of  the  other. 

The  word  "  rotation,"  as  employed  in  this  definition,  means 
the  amount  of  turning  effected,  and  not  the  process  of  turning. 

Def.  2. — For  convenience  the  lines  OP  and  OP',  which,  by 
their  difference  in  direction  form  the  angle,  are  called  the 
arms  of  the  angle,  and  the  point  O  where  the  arms  meet  is 
ih.^  vertex. 

Cor.  From  31°,  2,  an  angle  does  not  in  any  way  depend 
upon  the  lengths  of  its  arms,  but  only  upon  their  relative 
directions. 

33°.  Notation  of  Angles. — i.  The  symbol  z.  is  used  for  the 
word  "  angle." 

2.  When  two  segments  meet  at  a  vertex  the  angle  between 
them  may  be  denoted  by  a  single  letter  ^^^^^^B 

placed  at  the  vertex,  as  the  ^O  ,  or  by 
a  letter  with  or  without  an  arch  of  dots, 
as  /^S  ;  or  by  three  letters  of  which  the 
extreme  ones  denote  points  upon  the  arms  of  the  angle  and 
the  middle  one  denotes  the  vertex,  as  lAOB. 


l6  SYNTHETIC  GEOMETRY. 

3.  The  angle  between  two  lines,  when  the  vertex  is  not 
pictured,  or  not  referred  to,  is  expressed  by  l{L.  M),  or  LM, 
where  L  and  M  denote  the  lines  in  the  one-letter  notation 
(22°);  or  z.(AB,  CD),  where  AB  and  CD  denote  the  Hnes  in 
the  two-letter  notation. 

-Two  angles  are  equal  when  the  arms  of  the  one 
may  be  made  to  coincide  in  direction  respec- 
tively with  the  arms  of  the  other ;  or  when 
the  angles  are  described  by  the  same  rotation. 
Thus,  if,  when  C  is  placed  upon  O,  and 
O'A'  is  made  to  lie  along  OA,  O'B'  can  also 
be  made  to  lie  along  OB,  the  ^lA'O'B'  is  equal 
to  lAOB.    This  equality  is  symbolized  thus  : 
Z-A'0'B'=^A0B. 
Where  the  sign  =  is  to  be  interpreted  as 
indicating  the  possibility  of  coincidence  by  superposition. 

35".  Sum  and  Difference  of  Angles. — The  sutn  of  two 
angles  is  the  angle  described  by  a  radius  vector  which 
describes  the  two  angles,  or  their  equals,  in  succession. 

sp'  Thus  if  a  radius  vector  starts  from  co- 

incidence   with    OA    and    rotates    into 
direction   OP   it   describes    the    zJVOP. 
If  it  next  rotates  into  direction  OP'  it 
d  A  describes  the  /.POP'.     But  in  its  whole 

rotation  it  has  described  the  zAOP'.     Therefore, 

^AOP'  =  ^AOP-H^POP'. 
Similarly,  ^AOP  =  ^AOP'-^POP'. 

Z>i?/:— When  two  angles,  as  AOP  and  POP',  have  one  arm 
in  common  lying  between  the  remaining  arms,  the  angles  are 
adjacent  angles. 

36°.  Def. — A  radius  vector  which  starts  from  any  given 
direction  and  makes  a  complete  rotation  so  as  to  return  to  its 
original  direction  describes  a  circumangley  ox  perigon. 


RELATIONS   OF   TWO   LINES.— ANGLES.  1 7 

One-half  of  a  circumangle  is  a  straight  angle,  and  one- 
fourth  of  a  circumangle  is  a  right  angle. 

37°.   Theorem. — If  any  number  of  lines  meet  in  a  point, 
the  sum  of  all  the  adjacent  angles  formed  is  a  circumangle. 

OA,  OB,  OC,  ...,  OF  are  lines  meeting 
in  O.     Then 
^AOB  +  ^BOC-f-^COD  +  ...-f^FOA 

=  a  circumangle.  D. 
Proof. — A  radius  vector  which  starts 
from  coincidence  with  OA  and  rotates  into 
the  successive   directions,  OB,  OC,   ...,  1^  ^ 

OF,  OA  describes  in  succession  the  angles  AOB,  BOC,  ..., 
EOF,  FOA. 

But  in  its  complete  rotation  it  describes  a  circumangle  (36°). 
/-AOB +  ^BOC  +  ...  +  Z.FOA  =  a  circumangle.      q.e.d. 
Cor.  The  result  may  be  thus  stated  : — 

The  sum  of  all  the  adjacent  angles  about  a  point  in 
the  plane  is  a  circumangle. 

38°.   Theorem. — The  sum  of  all  the  adjacent  angles  on  one 
side  of  a  line,  and  about  a  point  in  the 
line  is  a  straight  angle. 

O  is  a  point  in  the  line  AB  ;  then 
^AOC  +  iLCOB  =  a  straight  angle. 

Proof. — Let  A  and  B  be  any  two  points 
in  the  line,  and  let  the  figure  formed  by  X*' 

AB  and  OC  be  revolved  about  AB  without  displacing  the 
points  A  and  B,  so  that  OC  may  come  into  a  position  OC. 
Then  (24°,  Cor.  2)  O  is  not  displaced  by  the  revolution, 
^AOC  =  ^AOC',  and  ^BOC  =  ^BOC'; 
.lAOC  +  :LBOC  =  £.AOC'  +  ^BOC', 
and  since  the  sum  of  the  four  angles  is  a  circumangle  {zT^^ 
therefore  the  sum  of  each  pair  is  a  straight  angle  (36°).   q.e.d. 
Cor.  I.  The  angle  between  the  opposite  directions  of  a  line 
is  a  straight  angle. 

B 


1 8  SYNTHETIC   GEOMETRY. 

Cor.  2.  If  a  radius  vector  be  rotated  until  its  direction  is 
reversed  it  describes  a  straight  angle.  And  conversely,  if  a 
radius  vector  describes  a  straight  angle  its  original  direction 
is  reversed. 

Thus,  if  OA  rotates  through  a  straight  angle  it  comes  into 
the  direction  OB.  And  conversely,  if  it  rotates  from  direction 
OA  to  direction  OB  it  describes  a  straight  angle. 

39°.  When  two  lines  L  and  M  cut  one 
another  four  angles  are  formed  about  the 
point  of  intersection,  any  one  of  which 
may  be  taken  to  be  the  angle  between 
the  lines. 

These  four  angles  consist  of  two  pairs  of  opposite  or 
vertical  angles,  viz.,  A,  A',  and  B,  B',  A  being  opposite  A', 
and  B  being  opposite  B'. 

40°.  Theorem. — The  opposite  angles  of  a  pair  formed  by 
two  intersecting  lines  are  equal  to  one  another. 

Proof.—          ^A  +  ^B  =  a  straight  angle  (38°) 

and                     Z.A'  +  A.B  =  a  straight  angle.  (38°) 

-lA=^A', 

and                              ^B  =  ^B'.  q.e.d. 

Def.  I. — Two  angles  which  together  make  up  a  straight 
angle  are  supplementary  to  one  another,  and  one  is  called  the 
supplement  of  the  other.  Thus,  A  is  the  supplement  of  B', 
and  B  of  A'. 

Cor.  If  aA  =  aB,  then  Z.A'  =  ^B=z.B',  and  all  four  angles 
are  equal,  and  each  is  a  right  angle  (36°). 

Therefore,  if  two  adjacent  angles  formed  by  two  intersect- 
ing lines  are  equal  to  each  other,  all  four  of  the  angles  so 
formed  are  equal  to  one  another,  and  each  is  a  right  angle. 

Def.  2. — When  two  intersecting  lines  form  a  right  angle  at 
their  point  of  intersection,  they  are  said  to  be  perpendicular 
to  one  another,  and  each  is  perpendicular  to  the  other. 


RELATIONS   OF   TWO   LINES. — ANGLES.  IQ 

Perpendicularity  is  denoted  by  the  symbol  X,  to  be  read 
"perpendicular  to"  or  "is  perpendicular  to." 

A  right  angle  is  denoted  by  the  symbol  ~\. 

The  symbol  ±  also  denotes  two  right  angles  or  a  straight 
angle. 

De/.  3. — When  two  angles  together  make  up  a  right  angle 
they  are  complementary  to  one  another,  and  each  is  the 
C07nplement  of  the  other. 

The  right  angle  is  the  simplest  of  all  angles,  for  when  two 
lines  form  an  angle  they  form  four  angles  equal  in  opposite 
pairs.  But  if  any  one  of  these  is  a  right  angle,  all  four  are 
right  angles. 

Perpendicularity  is  the  most  important  directional  relation 
in  the  applications  of  Geometry. 

Def.  4. — An  acute  angle  is  less  than  a  right  angle,  and  an 
obtuse  angle  is  greater  than  a  right  angle,  and  less  than  two 
right  angles. 

41°.  From  (36°)  we  have 

I  circumangle  =  2  straight  angles 
=4  right  angles. 
In  estimating  an  angle  numerically  it  may  be  expressed  in 
any  one  of  the  given  units. 

If  a  right  angle  be  taken  as  the  unit,  a  circumangle  is 
expressed  by  4,  i.e.  four  right  angles,  and  a  straight  angle 
by  2. 

Angles  less  than  a  right  angle  may  be  expressed,  approxi- 
mately at  least,  by  fractions,  or  as  fractional  parts  of  the 
right  angle. 

For  practical  purposes  the  right  angle  is  divided  into  90 
equal  parts  called  degrees  ;  each  degree  is  divided  into  60 
equal  parts  called  minutes  ;  and  each  minute  into  60  equal 
parts  called  seconds. 

Thus  an  angle  which  is  one-seventh  of  a  circumangle 
contains  fifty-one  degrees,  twenty-five  minutes,  and  forty-two 


20  SYNTHETIC   GEOMETRY. 

seconds  and  six-sevenths  of  a  second.     This  is  denoted  as 
follows  : —  51°  25'  42f". 

42°.   Theorem. — Through  a  given  point  in  a  line  only  one 
perpendicular  can  be  drawn  to  the  line. 

The  hne  OC  is  _L  AB,  and  OD  is  any 
other  line  through  O. 
Then  OD  is  not  _L  AB. 
Proof.— 'Y\i^  angles   BOC   and   COA 
are  each  right  angles  (40°,  Def.  2). 
Therefore  BOD  is  not  a  right  angle,  and  OD  is  not  ±  AB. 
But  OD  is  any  line  other  than  OC. 

Therefore  OC  is  the  only  perpendicular.  q.e.d. 

Def. — The  perpendicular  to  a  line-segment  through  its 
middle  point  is  the  right  bisector  of  the  segment. 

Since  a  segment  has  but  one  middle  point  (30°,  Ex.  3),  and 
since  but  one  perpendicular  can  be  drawn  to  the  segment 
through  that  point, 

.*.     a  Hite-segmeiit  has  but  one  right  bisector, 

43°.  Def. — The  lines  which  pass  through  the  vertex  of  an 
angle  and  make  equal  angles  with  the  arms,  are  the  bisectors 
of  the  angle.  The  one  which  lies  within  the  angle  is  the 
internal  bisector,  and  the  one  lying  without  is  the  external 
bisector. 

Let  AOC  be  a  given  angle ;  and 
■£     let  EOF  be  so  drawn  that  ^AOE 
=  Z.EOC. 

EF  is  the  internal  bisector  of  the 
angle  AOC. 
^  Also,  let  GOH  be  so  drawn  that 

D  ^COG  =  ^HOA. 

HG  is  the  external  bisector  of  the  angle  AOC. 

z.COG  =  ^HOA  (hyp.) 

and  z.HOA  =  z_GOB,  (40°) 

^COG=^GOB; 


RELATIONS   OF   TWO   LINES. — ANGLES.         21 

and  the  external  bisector  of  AOC  is  the  internal  bisector  of 
its  supplementary  angle,  COB,  and  vice  versa. 

The  reason  for  calling  GH  a  bisector  of  the  angle  AOC  is 
given  in  the  definition,  viz.,  GH  makes  equal  angles  with  the 
arms.  Also,  OA  and  OC  are  only  parts  of  indefinite  lines, 
whose  angle  of  intersection  may  be  taken  as  the  Z.AOC  or 
as  the  -lCOB. 

44°.  Just  as  in  23°  we  found  two  points  which  are  said 
to  divide  the  segment  in  the  same  manner,  so  we  may  find 
two  lines  dividing  a  given  angle  in  the  same  manner,  one 
dividing  it  internally,  and  the  other  externally. 

Thus,  if  OE  is  so  drawn  that  the  Zj\OE  is  double  the 
/_EOC,  some  line  OG  may  also  be  drawn  so  that  the  ^AOG 
is  double  the  Z.GOC. 

This  double  relation  in  the  division  of  a  segment  or  an 
angle  is  of  the  highest  importance  in  Geometry. 

45°.  Theorem. — The  bisectors  of  an  angle  are  perpendicular 
to  one  another. 

EF  and  GH  are  bisectors  of  the  Z-AOC ; 
then  EF  is  ±  GH. 

Proof.- 
and 

adding, 
But 


Exercises. 

1.  Three  lines  pass  through  a  common  point  and  divide  the 

plane  into  6  equal  angles.     Express  the  value  of  each 
angle  in  right  angles,  and  in  degrees. 

2.  OA  and  OB  make  an  angle  of  30°,  how  many  degrees  are 

there  in  the  angle  made  by  OA  and  the  external  bisector 
of  the  angle  AOB.^ 


LS.OC- 

=  i^OC,     V   OE 

is  a 

bisector ; 

LCOG: 

=|^COB,     V   OG 

is  a 

bisector ; 

Z.EOG: 

=  |^AOB. 

Z.AOB 

is  a  straight  angle, 

(38°, 

Cor.  i) 

^EOG 

is  a  right  angle. 

(36°) 

22  SYNTHETIC   GEOMETRY. 

3.  What  is  the  supplement  of  13°  27'  42"?     What  is  its 

complement  ? 

4.  Two  lines  make  an  angle  a  with  one  another,  and  the 

bisectors  of  the  angle  are  drawn,  and  again  the  bisectors 
of  the  angle  between  these  bisectors.  What  are  the 
angles  between  these  latter  lines  and  the  original  ones  ? 

5.  The  lines  L,  M  intersect  at  O,  and  through  O,  V  and  M' 

are  drawn  J_  respectively  to  L  and  M.  The  angle  be- 
tween L'  and  M'  is  equal  to  that  between  L  and  M. 


SECTION   III. 

THREE   OR  MORE   POINTS  AND  LINES. 
THE   TRIANGLE. 

46°.  Theorem. — Three  points  determine  at  most  three  lines ; 
and  three  lines  determine  at  most  three  points. 

Proof  I. — Since  (24°,  Cor.  2)  two  points  determine  one 
line,  three  points  detemiine  as  many  lines 
as  we  can  form  groups  from  three  points 
taken  two  and  two. 

Let  A,  B,   C  be  the  points  ;   the  groups 
are  AB,  BC,  and  CA. 

Therefore  three  points  determine  at  most  three  lines. 

2. — Since  (24°,  Cor.  3)  two  lines  determine  one  point,  three 
lines  determine  as  many  points  as  we  can  form  groups  from 
three  lines  taken  two  and  two. 

But  if  L,  M,  N  be  the  Hnes  the  groups  are  LM,  MN,  and 
NL. 

Therefore  three  lines  determine  by  their  intersections  at 
most  three  points. 


THREE  OR    MORE   POINTS   AND   LINES. 


23 


47°.  Theorem. — Four  points  determine  at  most  six  lines ;  and 
four  lines  determine  at  most  six  points. 

Proof.—-!.  Let  A,  B,  C,  D  be  the  four 
points.  The  groups  of  two  are  AB,  AC, 
AD,  BC,  BD,  and  CD  ;  or  six  in  all. 

Therefore  six  lines  at  most  are  deter- 
mined. 

2.  Let  L,  M,  N,  K  be  the  lines.  The 
groups  of  two  that  can  be  made  are  KL, 
KM,  KN,  LM,  LN,  and  MN  ;  or  six  in  all. 

Therefore  six  points  of  intersection  at 
most  are  determined. 

Cor.  In  the  first  case  the  six  lines  determined  pass  by 
threes  through  the  four  points.  And  in  the  second  case  the 
six  points  determined  lie  by  threes  upon  the  four  lines. 

This  reciprocality  of  property  is  very  suggestive,  and  in  the 
higher  Geometry  is  of  special  importance. 

Ex.  Show  that  5  points  determine  at  most  10  lines,  and  5 
lines  determine  at  most  10  points.  And  that  in  the  first  case 
the  lines  pass  by  fours  through  each  point ;  and  in  the  latter, 
the  points  lie  by  fours  on  each  line. 

48°.  Def. — A  triangle  is  the  figure  formed  by  three  lines 
and  the  determined  points,  or  by  three  points  and  the  deter- 
mined lines. 

The  points  are  the  vertices  of  the  triangle,  and  the  line- 
segments  which  have  the  points  as  end-points  are  the  sides. 

The  remaining  portions  of  the  determined  lines  are  usually 
spoken  of  as  the  "sides  produced."  But  in  many  cases 
generality  requires  us  to  extend  the  term  \ 

"  side  "  to  the  whole  line.  ^- 

Thus,  the  points  A,  B,  C  are  the 
vertices  of  the  triangle  ABC. 

The  segments  AB,  BC,  CA  are  the 
sides.  The  portions  AE,  BF,  CD,  etc., 
extending  outwards  as  far  as  required,  arc  the  sides  produced. 


24  SYNTHETIC   GEOMETRY. 

The  triangle  is  distinctive  in  being  the  rectilinear  figure  for 
which  a  given  number  of  lines  determines  the  same  number 
of  points,  or  vice  versa. 

Hence  when  the  three  points,  forming  the  vertices,  are 
given,  or  when  the  three  lines  or  line-segments  forming  the 
sides  are  given,  the  triangle  is  com- 
pletely given. 

This  is  not  the  case  with  a  rectilinear 
figure  having  any  number  of  vertices 
other  than  three. 

If  the  vertices  be  four  in  number, 

with  the  restriction  that  each  vertex  is 

determined  by  the  intersection  of  two 

sides,  any  one   of  the   figures    in  the 

^  margin  will  satisfy  the  conditions. 

Hence  the  giving  of  the  four  vertices  of  such  a  figure  is 

not  sufficient  to  completely  determine  the  figure. 

49°.  Def.—i.  The  angles  ABC,  BCA,  CAB  are  the  in- 
ternal  angles  of  the  triangle,  or  simply  the  angles  of  the 
triangle. 

2.  The  angle  DCB,  and  others  of  like  kind,  are  external 
angles  of  the  triangle. 

3.  In  relation  to  the  external  angle  DCB,  the  angle  BCA 
is  the  adjacent  interjial  angle,  while  the  angles  CAB  and 
ABC  are  opposite  iJiternal  angles. 

4.  Any  side  of  a  triangle  may  be  taken  as  its  base,  and  then 
the  angles  at  the  extremities  of  the  base  are  its  basal  angles, 
and  the  angle  opposite  the  base  is  the  vertical  angle.  The 
vertex  of  the  vertical  angle  is  the  vertex  of  the  triangle 
when  spoken  of  in  relation  to  the  base. 

50°.  Notation. — The  symbol  A  is  commonly  used  for  the 
word  triangle.  In  certain  cases,  which  are  always  readily 
apprehended,  it  denotes  the  area  of  the  triangle. 


THREE   OR   MORE   POINTS  AND   LINES.         25 

The  angles  of  the  triangle  are  denoted  usually  by  the 
capital  letters  A,  B,  C,  and  the  sides  opposite  by  the  cor- 
responding small  letters  a,  b^  c. 

51°.  Def. — When  two  figures  compared  by  superposition 
coincide  in  all  their  parts  and  become  virtually  but  one  figure 
they  are  said  to  be  congruent. 

Congruent  figures  are  distinguishable  from  one  another 
only  by  their  position  in  space  and  are  said  to  be  identically 
equal. 

Congruence  is  denoted  by  the  algebraic  symbol  of  identity, 
= ;  and  this  symbol  placed  between  two  figures  capable  of 
congruence  denotes  that  the  figures  are  congruent. 

Closed  figures,  like  triangles,  admit  of  comparison  in  two 
ways.  The  first  is  as  to  their  capability  of  perfect  coinci- 
dence ;  when  this  is  satisfied  the  figures  are  congruent.  The 
second  is  as  to  the  magnitude  or  extent  of  the  portions  of  the 
plane  enclosed  by  the  figures.  Equality  in  this  respect  is 
expressed  by  saying  that  the  figures  are  equal. 

When  only  one  kind  of  comparison  is  possible,  as  is  the 
case  with  line-segments  and  angles,  the  word  equal  is  used. 


CONGRUENCE   AMONGST   TRIANGLES. 

52°.  Theorem. — Two  triangles  are  congruent  when  two 
sides  and  the  included  angle  in  the  one  are  respectively  equal 
to  two  sides  and  the  included  angle  in  the  other. 

If  AB=A'B"|  the  triangles  y^B  /^B' 

BC  =  B'C  \  are    congru- 
and   z.B  =  z.B'  J  ent. 

Proof.— V\2.CQ  AABC  on  A  ~c      a^  C 

AA'B'C  so  that  B  coincides  with  B',  and  BA  lies  along  B'A'. 
z.B  =  /lB',  BC  lies  along  B'C,  (34°) 

and  V  AB  =  A'B'  and  BC  =  B'C'; 

A  coincides  with  A'  and  C  with  C,  (27") 


26  SYNTHETIC   GEOMETRY. 

and  .*.  AC  lies  along  A'C  ;  (24°,  Cor.  2) 

and  the  As  coinciding  in  all  their  parts  are  congruent.    (51°) 

q.e.d. 

Cor.  Since  two  congruent  triangles  can  be  made  to  coin- 
cide in  all  their  parts,  therefore — 

When  two  triangles  have  two  sides  and  the  included  angle 
in  the  one  respectively  equal  to  two  sides  and  the  included 
angle  in  the  other,  all  the  parts  in  the  one  are  respectively- 
equal  to  the  corresponding  parts  in  the  other. 

53°.  Theorem. — Every  point  upon  the  right  bisector  of  a 
segment  is  equidistant  from  the  end-points  of  the  segment 

AB  is  a  line-segment,  and  P  is  any  point 
on  its  right  bisector  PC.     Then  PA  =  PB. 

Proof.— In  the  As  APC  and  BPC, 

AC  =  CB,  (42°,  Def.) 

z_ACP  =  z_BCP,  (42°,  Def.) 

and  PC  is  common  to  both  As  ; 

AAPC  =  ABPC,  (52°) 

and.-.  PA  =  PB.  (52°,  Cor.)  ^.^.^/. 

Def.  I. — A  triangle  which  has  two  sides  equal  to  one  an- 
other is  an  isosceles  triangle. 

Thus  the  triangle  APB  is  isosceles. 

The  side  AB,  which  is  not  one  of  the  equal  sides,  is  called 
the  base. 

Cor.  I.  Since  the  AAPC=ABPC, 
^A=^B. 
Hence  the  basal  angles  of  an  isosceles  triangle  are  equal  to 
one  another. 

Cor.  2.  From  (52°,  Cor.),  ^APC  =  Z-BPC  ; 

Therefore  the  right  bisector  of  the  base  of  an  isosceles  tri- 
angle is  the  internal  bisector  of  the  vertical  angle.  And  since 
these  two  bisectors  are  one  and  the  same  line  the  converse  is 
true. 


THREE  OR   MORE   POINTS   AND   LINES. 


27 


Def.  2. — A  triangle  in  which  all  the  sides  are  equal  to  one 
another  is  an  equilateral  triangle. 

Cor.  3.  Since  an  equilateral  triangle  is  isosceles  with  re- 
spect to  each  side  as  base,  all  the  angles  of  an  equilateral 
triangle  are  equal  to  one  another  ;  or,  an  equilateral  triangle 
is  equiangular. 


54°.  Theorem. — Every  point  equidistant  from  the  end- 
points  of  a  line-segment  is  on  the  right  bisector  of  that 
segment.     (Converse  of  53°.) 

PA  =  PB.     Then  P  is  on  the  right  bisector 
ofAB.  Q^ 

Proof. — If  P  is  not  on  the  right  bisector  of 
AB,  let  the  right  bisector  cut  AP  in  Q. 

Then  QA=QB,  (53°) 

but  PA=PB,  (hyp.) 

QP  =  PB-QB, 
or  PB  =  QP  +  QB, 

which  is  not  true.  (25°,  Ax.) 

Therefore  the  right  bisector  of  AB  does  not  cut  AP  ;  and 
similarly  it  does  not  cut  BP  ;  therefore  it  passes  through  P, 
or  P  is  on  the  right  bisector.  q.e.d. 

This  form  of  proof  should  be  compared  with  that  of  Art. 
53°,  they  being  the  kinds  indicated  in  6°. 

This  latter  or  indirect  form  is  known  as  proof  by  reductio 
ad  absurdum  (leading  to  an  absurdity).  In  it  we  prove  the 
conclusion  of  the  theorem  to  be  true  by  showing  that  the 
acceptance  of  any  other  conclusion  leads  us  to  some  relation 
which  is  absurd  or  untrue. 


55°.  Def. — The  line-segment  from  a  vertex  of  a  triangle  to 
the  middle  of  the  opposite  side  is  a  median  of  the  triangle. 

Cor.  I.  Every  triangle  has  three  medians. 

Cor.  2.  The  median  to  the  base  of  an  isosceles  triangle  is 


28  SYNTHETIC  GEOMETRY. 

the  right  bisector  of  the  base,  and  the  internal  bisector  of  the 
vertical  angle.  (53°j  Cor.  2.) 

Cor.  3.  The  three  medians  of  an  equilateral  triangle  are 
the  three  right  bisectors  of  the  sides,  and  the  three  internal 
bisectors  of  the  angles. 

56°.  Theorem.— \i  two  angles  of  a  triangle  are  equal  to  one 
another,  the  triangle  is  isosceles,  and  the  equal  sides  are 
opposite  the  equal  angles.     (Converse  of  53°, 
Cor.  I.) 

z.PAB=^PBA,  then  PA  =  PB. 

Proof. — If  P  is  on  the  right  bisector  of 
AB,  PA=PB.  (53°) 

If  P  is  not  on  the  right  bisector,  let  AP 
A  fc  B  cut  the  right  bisector  in  Q. 

Then  QA=QB,  and  ^QAB=z.QBA.  (53°  and  Cor.  i) 

But         ^PBA  =  ^QAB;  (hyp.) 

z.PBA  =  ^QBA, 
which  is  not  true  unless  P  and  Q  coincide. 

Therefore  if  P  is  not  on  the  right  bisector  of  AB,  the 
Z.PAB  cannot  be  equal  to  the  Z-PBA. 
But  they  are  equal  by  hypothesis  ; 

P  is  on  the  right  bisector, 
and  PA=PB.  q.e.d. 

Cor.  If  all  the  angles  of  a  triangle  are  equal  to  one  another, 
all  the  sides  are  equal  to  one  another. 
Or,  an  equiangular  triangle  is  equilateral. 

57°.  From  53°  and  56°  it  follows  that  equality  amongst  the 
sides  of  a  triangle  is  accompanied  by  equality  amongst  the 
angles  opposite  these  sides,  and  conversely. 

Also,  that  if  no  two  sides  of  a  triangle  are  equal  to  one 
another,  then  no  two  angles  are  equal  to  one  another,  and 
conversely. 


THREE  OR   MORE   POINTS   AND   LINES. 


29 


Def. — A  triangle  which  has  no  two  sides  equal  to  one 
another  is  a  scalene  triangle. 

Hence  a  scalene  triangle  has  no  two  angles  equal  to  one 
another. 


58°.  Theorem. — If  two  triangles  have  the  three  sides  in  the 

one  respectively  equal  to  the  three  sides  in  the  other  the 

triangles  are  congruent.  b  b' 

If  A'B'  =  AB]the  As  ABC 

B'C'  =  BC  hand  A'B'C are 

C'A'  =  CA  J  congruent.      A< 

Proof.— iMXVi  the  AA'B'C 
over  and  place  A'  on  A,  and 
A'C  along  AC,  and  let  B'  fall  at  some  point  D. 

A'C'  =  AC,C  falls  ate,  (27°) 

and  AADC  is  the  AA'B'C  in  its  reversed  position. 

Since  AB  =  AD  and  CB  -  CD, 

A  and  C  are  on  the  right  bisector  of  BD,  and  AC  is  the  right 

bisector  of  BD.  (54°) 

Z.BAC =z.DAC  ;  (53°,  Cor.  2) 

and  the  As  BAC  and  DAC  are  congruent.  (52°) 

AABC  =  AA'B'C.  q.e.d. 


59°.  Theorem. — If  two  triangles  have  two  angles  and  the 
included  side  in  the  one  equal  respectively  to  two  angles  and 
the  included  side  in  the  other,  the  triangles  are  congruent. 

If      lM^lK'X 


and 


A'C'=ACJ 


the  As  ABC  and  A'B'C  are  congruent. 


A  c       A'  c' 

Proof. — Place  A'  on  A,  and  A'C  along  AC. 


30 


SYNTHETIC   GEOMETRY. 


Because        A'C'  =  AC,  C  coincides  with  C  ; 
and  V  lA.'=lI^,  A'B'  lies  along  AB  ; 

and  V  lC=lC,  CV>'  lies  along  CB  ; 

B'  coincides  with  B, 
and  the  triangles  are  congruent. 


(27°) 


(24°,  Cor.  3) 
q.e.d. 


60°.   Theorem. — An  external  angle  of  a  triangle  is  greater 
than  an  internal  opposite  angle. 

The  external  angle  BCD  is  greater 
than  the  internal  opposite  angle  ABC  or 
BAG. 

Proof. — Let  BF  be  a  median  produced 
until  FG  =  BF. 
Then  the  As  ABF  and  GGF  have 
BF  =  FG,  (construction) 

AF  =  FC,  •  (55°) 

and  z.BFA  =  ^GFC.  (40°) 

AABF=ACGF,  (52°) 

and  z.FCG  =  ^BAC.  (52°,  Cor.) 

But  /LACE  is  greater  than  Z.FCG. 

z.ACEis>^BAC. 
Similarly,  aBCD  is>^ABC, 

and  ^BGD=^ACE.  (40°) 

Therefore  the  ^s  BCD  and  ACE  are  each  greater  than  each 
of  the  Z.S  ABC  and  BAG.  q.e.d. 

61°.   Theorem. — Only  one  perpendicular  can  be  drawn  to  a 
line  from  a  point  not  on  the  line. 

Proof.— L^t  B  be  the  point  and  AD  the  line  ; 
and  let  BG  be  _L  to  AD,  and  BA  be  any  line 
other  than  BG. 

Then  aBCD  is>z.BAG,  (60°) 

^BAG  is  not  a  ~|, 
and  BA  is  not  ±  to  AD. 

But  BA  is  any  line  other  than  BG  ; 

BC  is  the  only  perpendicular  from  B  to  AD.       q.e.d 


A   0 


THREE   OR   MORE   POINTS   AND   LINES.  3 1 

Cor.  Combining  this  result  with  that  of  42°  we  have — 
Through  a  given  point  only  one  perpendicular  can  be  drawn 
to  a  given  line. 

62°.  Theorem.— Oi  any  two  unequal  sides  of  a  triangle 
and  the  opposite  angles — 

1.  The  greater  angle  is  opposite  the  longer  side. 

2.  The  longer  side  is  opposite  the  greater  angle. 

1.  BAis>BC;  b 
then                  Z.C  is  >  ^A. 

Proof.— L^t  BD  =  BC. 

Then  the  A^DC  is  isosceles,  **  "^^^ 

and  z_BDC  =  z.BCD.  (53°,  Cor.  i) 

But  z.BDCis>z.A,  (60°) 

and  ^BCAis>^BCD; 

/LBCAis>z.BAC; 
or,  Z.C  is>Z-A.  q.e.d. 

2.  lQ  is  >  ^A  ;  then  AB  is  >  BC. 

Proof. — From  the  Rule  of  Identity  (7°),  since  there  is  but 
one  longer  side  and  one  greater  angle,  and  since  it  is  shown 
(i)  that  the  greater  angle  is  opposite  the  longer  side,  therefore 
the  longer  side  is  opposite  the  greater  angle.  q.e.d. 

Cor.  I.  In  any  scalene  triangle  the  sides  being  unequal  to 
one  another,  the  greatest  angle  is  opposite  the  longest  side, 
and  the  longest  side  is  opposite  the  greatest  angle. 

Also,  the  shortest  side  is  opposite  the  smallest  angle,  and 
conversely. 

Hence  if  A,  B,  C  denote  the  angles,  and  a,  b,  c  the  sides 
respectively  opposite,  the  order  of  magnitude  of  A,  B,  C  is 
the  same  as  that  of  a.,  b,  c. 

63°.  Theorem. — Of  all  the  segments  between  a  given  point 
and  a  line  not  passing  through  the  point — 

1.  The  perpendicular  to  the  line  is  the  shortest. 

2.  Of  any  two  segments   the  one  which  meets  the  line 


I  SYNTHETIC   GEOMETRY. 

further  from  the  perpendicular  is  the  longer  ;  and  con- 
versely, the  longer  meets  the  line  further  from  the 
perpendicular  than  the  shorter  does. 
3.  Two,  and  only  two  segments  can  be  equal,  and  they  lie 
upon  opposite  sides  of  the  perpendicular. 

P  P  is  any  point  and  BC  a  line  not  pass- 

ing through  it,  and  PA  is  _L  to  BC. 

I.  PA  which   is  J_  to    BC    is   shorter 
than  any  segment  PB  which  is  not  ± 
"b  to  BC. 

Z-PAC  =  ^PAB=~1-  (hyp.) 

z.PACis>^PBC;  (60°) 
^PABis>^PBC, 

and  .-.                           PB  is  >  PA.  (62°,  2)  q.e.d. 

2.  AC  is  >  AB,  then  also  PC  is  >  PB. 

P;-^^— Since  AC  is  >  AB,  let  D  be  the  point  in  AC 
sothatAD  =  AB. 

Then  A  is  the  middle  point  of  BD,  and  PA  is  the  right 

bisector  of  BD.  (42°,  Def.) 

PD  =  PB  (53°) 

and  aPDB=^PBD.  (53°,  Cor  i) 

But  ^PDBis>^PCB; 

A.PBD  is>z.PCB, 
and  PCis>PB.  (62°,  2) 

The  converse  follows  from  the  Rule  of  Identity.  q.e.d. 

3.  Proof. — In  2  it  is  proved  that  PD  =  PB.  Therefore  two 
equal  segments  can  be  drawn  from  any  point  P  to  the  line 
BC  ;  and  these  lie  upon  opposite  sides  of  PA. 

No  other  segment  can  be  drawn  equal  to  PD  or  PB.  For 
it  must  lie  upon  the  same  side  of  the  perpendicular,  PA,  as 
one  of  them.  If  it  lies  further  from  the  perpendicular  than 
this  one  it  is  longer,  (2),  and  if  it  lies  nearer  the  perpendicular 
it  is  shorter.  Therefore  it  must  coincide  with  one  of  them 
and  is  not  a  third  line.  q.e.d 


THREE   OR   MORE   POINTS   AND    LINES.         33 

■l^^/'—The  length  of  the  perpendicular  segment  between 
any  point  and  a  line  is  the  distance  of  the  point  from  the 
line. 


64°.  Theorevt.~\i  two  triangles  have  two  angles  in  the  one 
respectively  equal  to  two  angles  in  the  other,  and  a  side 
opposite  an  equal  angle  in  each  equal,  the  triangles  are 
congruent. 

If    /-A'=^Althen      the      As 
^C'  =  aC  \h:^'C  and  ABC 
and   A'B'  =  AbJ  are  congruent. 

Proof.— V\2iQ^  A'   on    A,   and 
A'B'  along  AB. 

A'B'  =  AB,  B'  coincides  with  B. 

Also,  *.•  z.A'  =  ^A,  A'C  lies  along  AC. 

Now  if  O  does  not  coincide  with  C,  let  it  fall  at  some  other 
point,  D,  on  AC. 

Then,  •.'  AB  =  A'B',  AD  =  A'C,  and  lA==lM, 

.'.  AA'B'C  =  AABD,  (52°) 

and  ^ADB  =  z.C'.  (52°,  Cor.) 

But  lC-=lC,  (hyp.) 

.-.      ^ADB  =  ^C, 
which  is  not  true  unless  D  coincides  with  C. 

Therefore  C  must  fall  at  C,  and  the  As  ABC  and  A'B'C 
are  congruent. 

The  case  in  which  D  may  be  supposed  to  be  a  point  on 
AC  produced  is  not  necessary.  For  we  may  then  super- 
impose the  AABC  on  the  AA'B'C. 

65°.  Theorem. —  If  two  triangles  have  two  sides  in  the  one 
respectively  equal  to  two  sides  in  the  other,  and  an  angle 
opposite  an  equal  side  in  each  equal,  then — 

1.  If  the  equal  angles  be  opposite  the  longer  of  the  two 

sides  in  each,  the  triangles  are  congruent. 

2.  If  the  equal  angles  be  opposite  the  shorter  of  the  two 

C 


34 


SYNTHETIC  GEOMETRY. 


sides  in  each,  the  triangles  are  not  necessarily  con- 
gruent. 

A'B'  =  AB, 

B'C  =  BC, 

^A'=^A. 

I.  If    BCis>AB, 

AA'B'C  =  AABC. 

Proof.— Since    BC   is  >  AB,  therefore   B'C  is  >  A'B'. 
Place  A'  on  A  and  A'C  along  AC. 

^A'  =  ^A,  and  A'B' = AB, 

B'  coincides  with  B.  (34°,  27°) 

Let  BP  be  _L  AC  ; 

then  B'C  cannot  lie  between  BA  and  BP  (63°,  2),  but  must 
lie  on  the  same  side  as  BC ;  and  being  equal  to  BC,  the  lines 
B'C  and  BC  coincide  (63°,  3),  and  hence 

AA'B'C'  =  AABC.  q.eJ. 

2.  If  BC  is  <  AB,  the  As 
A'B'C  and  ABC  may  or  may 
not  be  congruent. 


-Since  AB  is  >  BC, 
(63°,  2) 


Proof.- 
PA  is  >  PC. 

Let  PD  =  PC, 

then  BD  =  BC. 

Now,  let  AA'B'C  be  superimposed  on  AABC  so  that  A' 
coincides  with  A,  B'  with  B,  and  A'C  lies  along  AC.  Then, 
since  we  are  not  given  the  length  of  A'C,  B'C  may  coincide 
with  BC,  and  the  As  A'B'C  and  ABC  be  congruent; 
or  B'C  may  coincide  with  BD,  and  the  triangles  A'B'C  and 
ABC  be  not  congruent.  q.e.d. 

Hence  when  two  triangles  have  two  sides  in  the  one 
respectively  equal  to  two  sides  in  the  other,  and  an  angle 
opposite  one  of  the  equal  sides  in  each  equal,  the  triangles 
are  not  necessarily  congruent  unless  some  other  relation 
exists  between  them. 


THREE  OR.  MORE  POINTS   AND   LINES.         35 

The  first  part  of  the  theorem  gives  one  of  the  suffici- 
ent relations.  Others  are  given  in  the  following  cor- 
ollaries. 

Cor.  I.  If  lC  is  a~J>  BC  and  BD  (2nd  Fig.)  coincide  along 
BP,  and  the  As  ABD  and  ABC  become  one  and  the  same. 
Hence  C  must  fall  at  C,  and  the  As  A'B'C  and  ABC  are 
congruent. 

Cor.  2.  The  Z.BDA  is  supplementary  to  BDC  and  therefore 
to  BCA.  And  •.•  lBDA  is  >  Z.BPA,  .'.  ^BDA  is  greater 
than  a  right  angle,  and  the  /.BCA  is  less  than  a  right 
angle. 

Hence  if,  in  addition  to  the  equalities  of  the  theorem,  the 
angles  C  and  C  are  both  equal  to,  or  both  greater  or  both 
less  than  a  right  angle,  the  triangles  are  congruent. 

De/.— Angles  which  are  both  greater  than,  or  both  equal  to, 
or  both  less  than  a  right  angle  are  said  to  be  0/  the  same 
affection. 

66°.  A  triangle  consists  of  six  parts,  three  sides  and  three 
angles.  When  two  triangles  are  congruent  all  the  parts  in 
the  one  are  respectively  equal  to  the  corresponding  parts  in 
the  other.  But  in  order  to  establish  the  congruence  of  two 
triangles  it  is  not  necessary  to  establish  independently  the 
respective  equality  of  all  the  parts ;  for,  as  has  now  been 
shown,  if  certain  of  the  corresponding  parts  be  equal  the 
equality  of  the  remaining  parts  and  hence  the  congruence  of 
the  triangles  follow  as  a  consequence.  Thus  it  is  sufficient 
that  two  sides  and  the  included  angle  in  one  triangle  shall  be 
respectively  equal  to  two  sides  and  the  included  angle  in 
another.  For,  if  we  are  given  these  parts,  we  are  given  con- 
sequentially all  the  parts  of  a  triangle,  since  every  triangle 
having  two  sides  and  the  included  angle  equal  respectively  to 
those  given  is  congruent  with  the  given  triangle. 

Hence  a  triangle  is  given  when  two  of  its  sides  and  the 
angle  between  them  are  given. 


36  SYNTHETIC   GEOMETRY. 

A  triangle  is  given  or  detertnined  by  its  elements  being 
given  according  to  the  following  table  : — 

1.  Three  sides, (58°) 

2.  Two  sides  and  the  included  angle, (52°) 

3.  Two  angles  and  the  included  side, (59°) 

4.  Two  angles  and  an  opposite  side, (64°) 

5.  Two  sides  and  the  angle  opposite  the  longer  side,  (65°) 

When  the  three  parts  given  are  two  sides  and  the  angle 
opposite  the  shorter  side,  two  triangles  satisfy  the  conditions, 
whereof  one  has  the  angle  opposite  the  longer  side  supple- 
mentary to  the  corresponding  angle  in  the  other. 

This  is  known  as  the  ambiguous  case  in  the  solution  of 
triangles. 

A  study  of  the  preceding  table  shows  that  a  triangle  is 
completely  given  when  any  three  of  its  six  parts  are  given, 
with  two  exceptions  : — 

(i)  The  three  angles  ; 

(2)  Two  sides  and  the  angle  opposite  the  shorter  of  the 
two  sides. 

67°.  Theorem. — If  two  triangles  have  two  sides  in  the  one 
respectively  equal  to  two  sides  in  the  other,  but  the  included 
angles  and  the  third  sides  unequal,  then 

1.  The  one  having  the  greater  included  angle  has   the 

greater  third  side. 

2.  Conversely,  the  one  having  the  greater  third  side  has  the 

greater  included  angle. 

^8'  A'B'  =  AB  and  B'C'  =  BC,  and 

I.        ^ABC  is  >  /lA'B'C, 
then  AC  is  >  A'C. 


Proof. — Let   A'  be  placed  upon 
A  and  A'B'  along  AB. 
E     c  "-^^,     Since  A'B' =  AB,  B' falls  on  B. 

"^  Let  C  fall  at  some  point  D. 

Then  ABD  is  A'B'C  in  its  new  position. 


THREE   OR   MORE   POINTS   AND   LINES.         37 

Let  BE  bisect  the  ^DBC  and  meet  AC  in  E. 

Join  DE. 

Then,  in  the  As  DBE  and  CBE, 

DB  =  BC,  (hyp.) 

^DBE  =  z.CBE,  (constr.) 

and  BE  is  common. 

ADBE  =  ACBE, 
and  DE  =  CE. 

But  AC  =  AE  +  EC  =  AE  +  ED, 

which  is  greater  than  AD. 

ACis>A'C'.  q.eJ. 

2.     AC  is  >  A'C,  then  ^ABC  is  greater  than  ^A'B'C 
Proof. — The  proof  of  this  follows  from  the  Rule  of  Identity. 

68°.  Theorem. — i.  Every  point  upon  a  bisector  of  an  angle 
is  equidistant  from  the  arms  of  the  angle. 

2.  Conversely,  every  point  equidistant  from  the  arms  of 
an  angle  is  on  one  of  the  bisectors  of  the  angle. 

1.  OP  and  OQ  are  bisectors  of  the  angle  AOB,  and  PA, 
PB  are  perpendiculars  from  P  upon 
the  arms.     Then 

PA=PB.  ''X  X^-^P 

Proof.— T\i^  As   POA  and    FOB   a' 
are  congruent,  since  they  have  two 
angles  and  an  opposite  side  equal  in 
each  (64°);    .-.  PA=PB. 

If  Q  be  a  point  on  the  bisector  OQ  it  is  shown  in  a  similar 
manner  that  the  perpendiculars  from  Q  upon  the  arms  of  the 
angle  AOB  are  equal.  q.e.d. 

2.  If  PA  is  i.  to  OA  and  PB  is  ±  to  OB,  and  PA  =  PB, 
then  PO  is  a  bisector  of  the  angle  AOB. 

Proof— 1\i^  As  POA  and  POB  are  congruent,  since  they 
have  two  sides  and  an  angle  opposite  the  longer  equal  in 
each  (65°,  i)  ;  .'.     z.POA  =  ^POB, 

and  PO  bisects  the  ^AOB. 


38  SYNTHETIC  GEOMETRY. 

Similarly,  if  the  perpendiculars  from  Q  upon  OA  ana  OB 
are  equal,  QO  bisects  the  i^BOA',  or  is  the  external  bisector 
of  the  ^AOB.  q.e.d. 


LOCUS. 

69°.  A  locus  is  the  figure  traced  by  a  variable  point,  which 
takes  all  possible  positions  subject  to  some  constraining 
condition. 

If  the  point  is  confined  to  the  plane  the  locus  is  one  or 
more  lines,  or  some  form  of  curve. 

Illustration. — In  the  practical  process  of  drawing  a  line  or 
curve  by  a  pencil,  the  point  of  the  pencil  becomes  a  variable 
(physical)  point,  and  the  line  or  curve  traced  is  its  locus. 

In  geometric  applications  the  point,  known  as  iht  generat- 
ing point,  moves  according  to  some  law. 

The  expression  of  this  law  in  the  Symbols  of  Algebra  is 
known  as  the  equation  to  the  locus. 

Cor.  I.  The  locus  of  a  point  in  the  plane,  equidistant  from 
the  end-points  of  a  given  line-segment,  is  the  right  bisector 
of  that  segment. 

This  appears  from  54°. 

Cor.  2.  The  locus  of  a  point  in  the  plane,  equidistant  from 
two  given  lines,  is  the  two  bisectors  of  the  angle  formed  by 
the  lines. 

This  appears  from  68°,  converse. 

Exercises. 

1.  How  many  lines  at  most  are  determined  by  5  points } 

by  6  points  ?  by  12  points  ? 

2.  How  many  points  at  most  are  determined  by  6  lines  "i  by 

12  lines  .? 

3.  How  many  points  are  determined  by  6  lines,  three  of  which 

pass  through  a  common  point  1 


THREE   OR   MORE    POINTS  AND   LINES.         39 

4.  How  many  angles  altogether  are  about  a  triangle?     How 

many  at  most  of  these  angles  are  different  in  magni- 
tude ?  What  is  the  least  number  of  angles  of  different 
magnitudes  about  a  triangle  ? 

5.  In  Fig.  of  53°,  if  Q  be  any  point  on  PC,  APAQ=APBQ. 

6.  In  Fig.  of  53°,  if  the  APCB  be  revolved  about  PC  as  an 

axis,  it  will  become  coincident  with  APCA. 

7.  The  medians  to  the  sides  of  an  isosceles  triangle  are 

equal  to  one  another. 

8.  Prove  58°  from  the  axiom  "  a  straight  line  is  the  shortest 

distance  between  two  given  points." 

9.  Show  from  60°  that  a  triangle  cannot  have  two  of  its 

angles  right  angles. 

10.  If  a  triangle  has  a  right  angle,  the  side  opposite  that  angle 

is  greater  than  either  of  the  other  sides. 

11.  What  is  the  locus  of  a  point  equidistant  from  two  sides 

of  a  triangle  ? 

12.  Find  the  locus  of  a  point  which  is  twice  as  far  from  one 

of  two  given  lines  as  from  the  other. 

13.  Find  the  locus  of  a  point  equidistant  from  a  given  line 

and  a  given  point. 


SECTION   IV. 

PARALLELS,   ETC 

70°.  Def. — Two  lines,  in  the  same  plane,  which  do  not 
intersect  at  any  finite  point  are  parallel. 

Next  to  perpendicularity,  parallelism  is  the  most  important 
directional  relation.  It  is  denoted  by  the  symbol  ||,  which  is 
to  be  read  "  parallel  to  "  or  "  is  parallel  to  "  as  occasion  may 
require. 

The  idea  of  parallelism  is  identical  with  that  of  sameness 


40  SYNTHETIC   GEOMETRY. 

of  direction.     Two  line-segments  may  differ  in  length  or  in 
direction  or  in  both. 

If,  irrespective  of  direction,  they  have  the  same  length, 
they  are  equal ;  if,  irrespective  of  length,  they  have  the  same 
direction,  they  are  parallel ;  and  if  both  length  and  direction 
are  the  same  they  are  equal  and  parallel.  Now  when  two 
segments  are  equal  one  may  be  made  to  coincide  with  the 
other  by  superposition  without  change  of  length,  whether 
change  of  direction  is  required  or  not.  So  when  they 
are  parallel  one  may  be  made  to  coincide  with  the  other 
without  change  of  direction,  whether  change  of  length  is 
required  or  not. 

Axiom. — Through  a  given  point  only  one  line  can  be 
drawn  parallel  to  a  given  line. 

This  axiom  may  be  derived  directly  from  24°. 

71°.  Theorem. — Two  lines  which  are  perpendicular  to  the 
same  line  are  parallel. 

L  and  M  are  both  ±  to  N, 
then  L  is  ||  to  M. 

Proof. — If  L  and  M  meet  at  any  point,  two 
.    perpendiculars  are  drawn  from  that  point  to 
the  line  N. 
But  this  is  impossible  (61°). 

Therefore  L  and  M  do  not  meet,  or  they  are  parallel. 
Cor.  All  lines  perpendicular  to  the  same  line  are  parallel 
to  one  another. 

72°.  Theorem. — Two  lines  which  are  parallel  are  perpen- 
dicular to  the  same  line,  or  they  have  a  common  perpendicular. 
(Converse  of  71°.) 

L  is  II  to  M,  and  L  is  ±  to  N  ; 
then  M  is  JL  to  N. 

Proof.— \i  M  is  not  _L  to  N,  through  any  point  P  in  M,  let 
K  be  ±  to  N. 


PARALLELS,  ETC.  4 1 

Then  K  is  ||  to  L.  (71°) 

But  M  is  II  to  L.  (hyp.) 

Therefore     K  and  M  are  both  jj  to  L, 
which  is  impossible  unless  K  and  M  coincide.  (70°,  Ax.) 

Therefore  L  and  M  are  both  J_  to  N, 
or  N  is  a  common  perpendicular. 

'j'^.  Def. — A  line  which  crosses   two  or  more   lines    of 
any  system  of  lines  is  a  transversal.  E/ 

Thus  EF  is  a  transversal  to  the  lines   ^  ^/        g 

AB  and  CD. 

In  general,  the  angles  formed  by    ^ 
a  transversal  to   any   two    lines  are  ^/h 

distinguished  as  follows —  /f 

a  and  e^  c  and^,  b  and/,  d  and  h  are  pairs  of  corresponding 
angles. 

c  and  f^  e  and  d  are  pairs  of  alternate  angles. 

c  and  <?,  ^and/are  pairs  of  interadjacent  angles. 

74°.  When  a  transversal  crosses  parallel  lines — 

1.  The  alternate  angles  are  equal  in  pairs. 

2.  The  corresponding  angles  are  equal  in  pairs. 

3.  The  sum  of  a  pair  of  interadjacent  angles  is  a  straight 

angle.                                                                     g^ 
AB  is  II  to  CD  and  EF  is  a  transversal,  a p    / b_ 

1.  ^AEF  =  ^EFD. 
Proof. — Through  O,  the  middle  point 

of  EF,  draw  PQ  a  common  _L  to  AB 

and  CD.  (72°) 

Then  AOPE  =  AOQF ;  (64°) 

zAEF  =  Z.EFD.  q.e.d. 

Similarly  the  remaining  alternate  angles  are  equal. 

2.  ^AEG  =  ^CFE,  etc. 

Proof.—        Z.AE  G  =  supplement  of  zJVEF,      (40°,  Def.  i) 
and  z_CFE  =  supplement  of  -lEFD. 


42  SYNTHETIC   GEOMETRY. 

But  zJVEF=^EFD ;  (74°,  i) 

^AEG  =  ^CFE.  q.e.d. 

Similarly  the  other  corresponding  angles  are  equal  in  pairs. 

3.  z_AEF  +  ^CFE=±. 

Proof.—  ^AEF=^EFD,  (74°,  0 

and  £.CFE  +  ^EFD=_L;  (38°) 

zAEF+^CFE=±.  q.e.d. 

Cor.  It  is  seen  from  the  theorem  that  the  equality  of  a  pair 
of  alternate  angles  determines  the  equality  in  pairs  of  corre- 
sponding angles,  and  also  determines  that  the  sum  of  a  pair 
of  interadjacent  angles  shall  be  a  straight  angle.  So  that  the 
truth  of  any  one  of  the  statements  i,  2,  3  determines  the  truth 
of  the  other  two,  and  hence  if  any  one  of  the  statements  be 
proved  the  others  are  indirectly  proved  also. 

75°.  Theorem. — If  a  transversal  to  two  lines  makes  a  pair 
of  alternate  angles  equal,  the  two  lines  are  parallel.  (Con- 
verse of  74°  in  part.) 

If  ^AEF  =  Z-EFD,  AB  and  CD  are  parallel. 

/'/w/— Draw  PQ  as  in  74°,  ±  to  AB, 
AOPE  =  AOQF;  (59°) 

.-.        :lOPE=^OQF=~l 
Q  and.-.        AB  is  |1  to  CD.         {7i°)q.e.d. 

Cor.  It  follows  from  74°  Cor.  that  if  a  pair  of  corresponding 
angles  are  equal  to  one  another,  or  if  the  sum  of  a  pair  of 
interadjacent  angles  is  a  straight  angle,  the  two  lines  are 
parallel. 

76°.  Theoreju. — The  sum  of  the  internal  angles  of  a  tri- 
angle is  a  straight  angle. 

ABC  is  a  A  ; 
the^A  +  ^B  +  ^C=±. 

A  c  D  Proof.— l.^X  CE  be  ||  to  AB,  and 

D  be  any  point  on  AC  produced. 


A 

P       / 

B 

0 

/^ 

c 

f/ 

0 

PARALLELS,   ETC.  43 

Then  BC  is  a  transversal  to  the  parallels  AB  and  CE  ; 

zlABC=a.BCE.  (74°,  i) 

Also,  AC  is  a  transversal  to  the  same  parallels  ; 

z.BAC=^ECD.  (74°,  2) 

^BC  +  ^BAC=^BCD 

=  supplement  of  Z.BCA. 
lA  +  lB  +  lC=±.  q.e.d. 

Cor.  An  external  angle  of  any  triangle  is  equal  to  the  sum 
of  the  opposite  internal  angles.  (49°,  3) 

For  ^BCD=^A+Z-B. 

T]".  From  the  property  that  the  sum  of  the  three  angles  of 
any  triangle  is  a  straight  angle,  and  therefore  constant,  we 
deduce  the  following — 

1.  When  two  angles  of  a  triangle  are  given  the  third  is 

given  also  ;  so  that  the  giving  of  the  third  furnishes 
no  new  information. 

2.  As  two  parts  of  a  triangle  are  not  sufficient  to  deter- 

mine it,  a  triangle  is  not  determined  by  its  three  angles, 
and  hence  one  side,  at  least,  must  be  given  (66°,  i). 

3.  The  magnitude  of  any  particular  angle  of  a  triangle  does 

not  depend  upon  the  size  of  the  triangle,  but  upon 
the  form  only,  i.e.,  upon  the  relations  amongst  the 
sides. 

4.  Two  triangles  may  have  their  angles  respectively  equal 

and  not  be  congruent.  But  such  triangles  have  the 
same  form  and  are  said  to  be  similar. 

5.  A  triangle  can  have  but  one  obtuse  angle ;  it  is  then 

called  an  obtuse-angled  triangle. 
A  triangle  can  have  but  one  right  angle,  when  it  is  called 

a  right-angled  triangle. 
All  other  triangles  are  called  acute-angled  triangles,  and 

have  three  acute  angles. 

6.  The  acute  angles  in  a  right-angled  triangle  are  comple- 

mentary to  one  another. 


44 


SYNTHETIC  GEOMETRY. 


78°.  Theorem. — If  a  line  cuts  a  given  line  it  cuts  every 
parallel  to  the  given  line. 

Let  L  cut  M,  and  let  N  be  any  parallel 
to  M.     Then  L  cuts  N. 


Proof. — If  L  does  not  cut  N  it  is  ||  to  N. 
But  M  is  II  to  N.  Therefore  through  the  same  point  P  two 
lines  L  and  M  pass  which  are  both  ||  to  N. 

But  this  is  impossible  ;  (70°,  Ax.) 

L  cuts  N. 
And  N  is  any  line  ||  to  M. 

L  cuts  every  line  ||  to  M.  q.e.d. 

79°.   Theorem. — If  a  transversal  to  two  lines  makes  the 
sum  of  a  pair  of  interadjacent  angles  less  than  a  straight 
angle,  the  two  lines  meet  upon  that  side 
of  the  transversal  upon  which  these  inter- 
adjacent angles  lie. 

GH  is  a  transversal  to  AB  and  CD, 
and  ^BEF-|-^EFD<±. 

Then  AB  and  CD  meet  towards  B  and  D. 
Proof. — Let  LK  pass  through  E  making  Z.KEF  =  Z.EFC. 
Then  LKis||toCD. 

But  AB  cuts  LK  in  E, 

it  cuts  CD.  (78°) 

Again,  •.*  EB  lies  between  the  parallels,  and  AE  does  not, 

the  point  where  AB  meets  CD  must  be  on  the  side  BD  of 

the  transversal.  q.e.d. 

Cor.  Two  lines,  which  are  respectively  perpendicular  to 

two  intersecting  lines,  intersect  at  some  finite  point. 

80°.  Def — I.  A   closed   figure    having    four 
lines  as  sides  is  in  general  called  a  quadrangle 
'c    or  quadrilateral. 
in  Thus  ABCD  is  a  quadrangle. 

2.  The  line-segments  AC   and   BD    which  join  opposite 
vertices  are  the  diagonals  of  the  quadrangle. 


PARALLELS,   ETC.  45 

3.  The  quadrangle  formed  when  two  parallel  lines  intersect 
two  other  parallel  lines  is  a  para  lie  logra7?t,  and  is  usually- 
denoted  by  the  symbol  1      7. 

81°.   Theorem. — In  any  parallelogram — 

1.  The  opposite  sides  are  equal  to  one  another. 

2.  The  opposite  internal  angles  are  equal  to  one  another. 

3.  The  diagonals  bisect  one  another.  A_ 
AB  is  II  to  CD,  and  AC  is  ||  to  BD,  and 


AD  and  BC  are  diagonals.  Z--'''^^^ 

1.  Then  AB  =  CDandAC  =  BD.  c  d 
Proof. — •.•  AD  is  a  transversal  to  the  parallels  AB  and  CD, 

^CDA=^DAB.  (74°,  i) 

and  •.'  AD  is  a  transversal  to  the  parallels  AC  and  BD, 

^CAD=^ADB.  (74°,  i) 

Hence,  ACAD  =  ABDA.  (59°) 

AB  =  CD  and  AC  =  BD.  q.e.d. 

2.  iLCAB  =  z.BDCand^ACD  =  ^DBA. 

Proof.— It  is  shown  in  i  that  z.CAD=^DB  and  Z.BAD 
=  ^DC; 

.*.  by  adding  equals  to  equals, 

^CAB  =  ^CDB. 
Similarly,  ^ACD  =  ^ABD.  q.e.d. 

3.  AO  =  OD  and  BO  =  OC. 

Proof  -The  AAO  C  =  AD  O  B  ;  (59°) 

AO  =  OD  andBO  =  OC.  q.e.d. 

82°.  Def  I.— A  parallelogram  which  has  two  adjacent  sides 
equal  is  a  rhombus.  B^ 

Cor.  I.  Since  AB  =  BC  (hyp.) 

=  DC(8i°,  i)=AD.     A< 

Therefore  a  rhombus  has  all  its  sides 
equal  to  one  another. 

Cor.  2.  Since  AC  is  the  right  bisector  of  BD, 
and  BD  the  right  bisector  of  AC, 


46  SYNTHETIC  GEOMETRY. 

Therefore  the  diagonals  of  a  rhombus  bisect  one  another 
at  right  angles. 

Def.  2. — A  parallelogram  which  has  one  right  angle  is  a 
rectangle^  and  is  denoted  by  the  symbol  en. 

Cor.  3.  Since  the  opposite  angle  is  a  ~~1,  (81°,  2) 

and  the  adjacent  angle  is  a  ""],  (74°)  3) 

Therefore  a  rectangle  has  all  its  angles  right  angles. 

Cor.  4.  The  diagonals  of  a  rectangle  are  equal  to  one 
another. 

Def.  3. — A  rectangle  with  two  adjacent  sides  equal  is  a 
square^  denoted  by  the  symbol  □. 

Cor.  5.  Since  the  square  is  a  particular  form  of  the  rhombus 
and  a  particular  form  of  the  rectangle, 

Therefore  all  the  sides  of  a  square  are  equal  to  one  another; 
all  the  angles  of  a  square  are  right  angles  ;  and  the  diagonals 
of  a  square  are  equal,  and  bisect  each  other  at  right  angles. 

84°.  Theorem. — If  three  parallel  lines  intercept  equal  seg- 
ments upon  any  one  transversal  they  do  so  upon  every 
transversal. 

AE  is  a  transversal  to  the  three  parallels 
AB,  CD,  and  EF,  so  that  AC  =  CE,  and 
BF  is  any  other  transversal.  Then  BI) 
=  DF. 

Proof. — Let  GDH  passing  through  D 
be  II  to  AE. 

Then  AGDC  and  CDHE  are  Z=7s.  (80°,  3) 

GD=AC  =  CE  =  DH.  (81°,  i) 

Also,            ^GBD  =^DFH,  •.'  AG  is  ||  to  EF,  (74°,  i) 

and                  ^BDG  =  ^FDH  ;  (40°) 

ABDG  =  AFDH,  (64°) 

and                        BD  =  DF.  q.e.d. 

Def— The  figure  ABFE  is  a  trapezoid. 

Therefore  a  trapezoid  is  a  quadrangle  having  only  two 


PARALLELS,   ETC. 


47 


sides  parallel.     The  parallel  sides  are  the  major  and  minor 
bases  of  the  figure. 


Cor.  I.  Since 


and 


2CD=AG  +  EH, 

=AB+BG+EF-HF 
BG=HF; 
CD  =  |(AB  +  EF). 
Or,  the  line-segment  joining  the  middle  points  of  the  non- 
parallel  sides  of  a  trapezoid  is  equal  to  one-half  the  sum  of 
the  parallel  sides. 

Cor.  2.  When  the  transversals  meet  upon 
one  of  the  extreme  parallels,  the  figure 
AEF'  becomes  a  A  and  CD'  becomes  a 
line  passing  through  the  middle  points  of 
the  sides  AE  and  AF',  and  parallel  to  the 
base  EF'. 

Therefore,  i,  the  line  through  the  middle  point  of  one  side 
of  a  triangle,  parallel  to  a  second  side,  bisects  the  third  side. 

And,  2,  the  line  through  the  middle  points  of  two  sides  of 
a  triangle  is  parallel  to  the  third  side. 

85°.  Theorem. — The  three  medians    of   a    triangle   pass 
through  a  common  point. 

CF  and  AD  are  medians  intersecting  in  O. 

Then  BO  is  the  median  to  AC. 

Proof.— L^\.  BO  cut  AC  in  E,  and  let 
AG  II  to  FC  meet  BO  in  G.    Join  CG. 

Then,  BAG  is  a  A  and  FO  passes  through 
the  middle  of  AB  and  is  ||  to  AG, 

.-.    O  is  the  middle  of  BG.      (84^  Cor  2) 

Again,  DO  passes  through  the  middle 
points  of  two  sides  of  the  ACBG, 

CG  is  II  to  AO  or  OD  ; 
AOCG  is  a  E=7, 
and  AE  =  EC; 

BO  is  the  median  to  AC. 


(84°,  Cor.  2) 

(81°,  3) 
q.e.d. 


48  SYNTHETIC   GEOMETRY. 

Def. — When  three  or  more  lines  meet  in  a  point  they  are 
said  to  be  concurrent. 
Therefore  the  three  medians  of  a  triangle  are  concurrent. 

Def.  2. — The  point  of  concurrence,  O,  of  the  medians  of  a 
triangle  is  the  centi'oid  oi  the  triangle. 

Cor.  Since  O  is  the  middle  point  of  BG,  and  E  is  the  middle 
point  of  OG,  (8i°,  3) 

OE  =  |OB, 
=  1EB. 
Therefore  the  centroid  of  a  triangle  divides  each  median 
at  two-thirds  of  its  length  from  its  vertex. 

86°.   Theorem. — The  three  right  bisectors  of  the  sides  of  a 
B  triangle  are  concurrent. 

/  Proof. — Let   L    and    N   be    the    right 

bisectors  of  BC  and  AB  respectively. 
Then  L  and  N  meet  in  some  point  O. 
A  ^  (79°,  Cor.) 

Since  L  is  the  right  bisector  of  BC,  and  N  of  AB,  O  is 
equidistant  from  B  and  C,  and  is  also  equidistant  from  A 
and  B.  (53°) 

Therefore  O  is  equidistant  from  A  and  C,  and  is  on  the 
right  bisector  of  AC.  (54°) 

Therefore  the  three  right  bisectors  meet  at  O.  q.e.d. 

Cor.  Since  two  lines  L  and  N  can  meet  in  only  one  point 
(24°,  Cor.  3),  O  is  the  only  point  in  the  plane  equidistant 
from  A,  B,  and  C. 

Therefore  only  one  finite  point  exists  in  the  plane  equi- 
distant from  three  given  points  in  the  plane. 

Def. — The  point  O,  for  reasons  given  hereafter,  is  called 
the  ctrcumce7itre  of  the  triangle  ABC. 

^j".  Def. — The  line  through  a  vertex  of  a  triangle  per- 
pendicular to  the  opposite  side  is  the  perpendicular  to  that 
side,  and  the  part  of  that  line  intercepted  within  the  triangle 
is  the  altitude  to  that  side. 


PARALLELS,    ETC.  49 

Where  no  reference  to  length  is  made  the  word  altitude  is 
often  employed  to  denote  the  indefinite  line  forming  the 
perpendicular. 

Hence  a  triangle  has  three  altitudes,  one  to  each  side. 

88°.  Theorem. — The  three  altitudes  of  a  triangle  are  con- 
current. 

Proof.— Let  ABC  be  a  triangle. 
Complete  the  zz::7s,  ACBF, 
ABDC,  and  ABCE. 

Then   •.•      FB  is  1|  to  AC, 
and  BD  is  ||  to  AC, 

FBD  is  one  line,   (70°,  Ax.) 
and  FB  =  BD. 

Similarly,     DCE  is  one  line  and  DC  =  CE, 
and  EAF  is  one  line  and  EA  =  AF. 

Now,  •••  AC  is  II  to  FD,  the  altitude  to  AC  is  J_  to  FD  and 
passes  through  B  the  middle  point  of  FD.  (72°) 

Therefore  the  altitude  to  AC  is  the  right  bisector  of  FD, 
and  similarly  the  altitudes  to  AB  and  BC  are  the  right  bisec- 
tors of  DE  and  EF  respectively. 

But  the  right  bisectors  of  the  sides  of  the  ADEF  are 
concurrent  (86°),  therefore  the  altitudes  of  the  AABC  are 
concurrent.  q.e.d. 

Def, — The  point  of  concurrence  of  the  altitudes  of  a  tri- 
angle is  the  orthocentre  of  the  triangle. 

Cor.  I.  If  a  triangle  is  acute-angled  {j']°^  5),  the  circum- 
centre  and  orthocentre  both  lie  within  the  triangle. 

2.  If  a  triangle  is  obtuse-angled,  the  circumcentre  and 
orthocentre  both  lie  without  the  triangle. 

3.  If  a  triangle  is  right-angled,  the  circumcentre  is  at  the 
middle  point  of  the  side  opposite  the  right  angle,  and  the 
orthocentre  is  the  right-angled  vertex. 

Def. — The  side  of  a  right-angled  triangle  opposite  the  right 
angle  is  called  the  hypothenuse. 

D 


50 


SYNTHETIC   GEOMETRY. 


89°.  The  definition  of  80°  admits  of  three  different  figures, 
viz. : — 

I.  The  normal  quadrangle  (i)  in  which  each  of  the  in- 
ternal angles  is  less  than  a  straight  angle.  When  not 
(i)  (2)  (3) 


otherwise  qualified  the  term  quadrangle  will  mean  this 
figure. 

2.  The  quadrangle  (2)  in  which  one  of  the  internal  angles, 
as  at  D,  is  greater  than  a  straight  angle.  Such  an  angle  in  a 
closed  figure  is  called  a  re-entrant  angle.  We  will  call  this 
an  inverted  quadrangle. 

3.  The  quadrangle  (3)  in  which  two  of  the  sides  cross  one 
another.     This  will  be  called  a  crossed  quadrangle. 

In  each  figure  AC  and  BD  are  the  diagonals  ^  so  that  both 
diagonals  are  within  in  the  normal  quadrangle,  one  is  within 
and  one  without  in  the  inverted  quadrangle,  and  both  are 
without  in  the  crossed  quadrangle. 

The  general  properties  of  the  quadrangle  are  common  to 
all  three  forms,  these  forms  being  only  variations  of  a  more 
general  figure  to  be  described  hereafter. 

90°.  Theorem. — The  sum  of  the  internal  angles  of  a  quad- 
rangle is  four  right  angles,  or  a  circumangle. 

Proof. — The  angles  of  the  two  As  ABD  and  CBD  make 
up  the  internal  angles  of  the  quadrangle. 

But  these  are  ±  +  _L ;  (76°) 

therefore  the  internal  angles  of  the  quadrangle  are  together 
equal  to  four  right  angles.  q.e.d. 

Cor.  This  theorem  applies  to  the  inverted  quadrangle  as  is 
readily  seen. 


PARALLELS,   ETC.  5 1 

91°.  Theorem. — If  two  lines  be  respectively  perpendicular 
to  two  other  lines,  the  angle  between  the  first  two  is  equal 
or  supplementary  to  the  angle  between  the  last  two. 

BC  is  _L  to  AB 
and  CD  is  _L  to  AD. 

Then  ^(BC .  CD)  is  equal  or  supplemen- 
tary to  4AB .  AD). 

Proof. — ABCD  is  a  quadrangle,  and  the 
Ls>  at  B  and  D  are  right  angles  ; 

^BAD+^BCD  =  J_, 
or  Z.BCD  is  supplementary  to  Z.BAD. 

But  /.BCD  is  supplementary  to  ^ECD  ; 

and  the  ^(BC .  CD)  is  either  the  angle  BCD  or  DCE.      (39°) 
Z.(BC .  CD)  is  =  or  supplementary  to  ^BAD.      q.e.d. 

Exercises. 

1.  ABC  is  a  A,  and  A',  B',  C  are  the  vertices  of  equilateral 

As  described  outwards  upon  the  sides  BC,  CA,  and 
AB  respectively.     Then  AA'=BB'  =  CC'.     (Use  52°.) 

2.  Is  Ex.   I   true  when   the   equilateral  As   are  described 

"  inwardly  "  or  upon  the  other  sides  of  their  bases  ? 

3.  Two  lines  which  are  parallel  to  the  same  line  are  parallel 

to  one  another. 

4.  L'  and  M'  are  two  lines  respectively  parallel  to  L  and  M. 

Thez.(L'.M')=^(L.M). 

5.  On  a  given  line  only  two  points  can  be  equidistant  from 

a  given  point.     How  are  they  situated  with  respect  to 
the  perpendicular  from  the  given  point  1 

6.  Any  side  of  a  A  is  greater  than  the  difference  between 

the  other  two  sides. 

7.  The  sum  of  the  segments  from  any  point  within  a  A  to 

the  three  vertices  is  less  than  the  perimeter  of  the  A 

8.  ABC  is  a  A  and  P  is  a  point  within  on  the  bisector  of 

z_A.     Then  the  difference  between  PB  and  PC  is  less 
than  that  between  AB  and  AC,  unless  the  A  is  isosceles. 


52  SYNTHETIC  GEOMETRY. 

9.  Is  Ex.  8  true  when  the  point  P  is  without  the  A?  but  on 
the  same  bisector  ? 

10.  Examine  Ex.  8  when  P  is  on  the  external  bisector  of  A, 

and  modify  the  wording  of  the  exercise  accordingly. 

11.  CE  and  CF  are  bisectors  of  the  angle  between  AB  and 

CD,  and  EF  is  parallel  to  AB.  Show  that  EF  is 
bisected  by  CD. 

12.  If  the  middle  points  of  the  sides  of  a  A  be  joined  two 

and  two,  the  A  is  divided  into  four  congruent  As. 

13.  From  any  point  in  a  side  of  an  equilateral  A  lines  are 

drawn  parallel  to  the  other  sides.  The  perimeter  of 
the  I     7  so  formed  is  equal  to  twice  a  side  of  the  A- 

14.  Examine   Ex.    13   when    the    point   is  on   a   side    pro- 

duced. 

15.  The  internal  bisector  of  one  angle  of  a  A  and  the  ex- 

ternal bisector  of  another  angle  meet  at  an  angle  which 
is  equal  to  one-half  the  third  angle  of  the  A 

16.  O  is  the  orthocentre  of  the  AABC.     Express  the  angles 

AOB,  BOC,  and  COA  in  terms  of  the  angles  A,  B, 
and  C. 

17.  P  is  the  circumcentre  of  the  AABC.     Express  the  angles 

APB,  BPC,  and  CPA  in  terms  of  the  angles  A,  B, 
and  C. 

1 8.  The  joins  of  the  middle  points  of  the  opposite  sides  of 

any  quadrangle  bisect  one  another. 
T9.  The  median  to  the  hypothenuse  of  a  right-angled  triangle 
is  equal  to  one-half  the  hypothenuse. 

20.  If  one  diagonal  of  a  1      7  be  equal  to  a  side  of  the  figure, 

the  other  diagonal  is  greater  than  any  side. 

21.  If  any  point  other  than  the  point  of  intersection  of  the 

diagonals  be  taken  in  a  quadrangle,  the  sum  of  the 
line-segments  joining  it  with  the  vertices  is  greater 
than  the  sum  of  the  diagonals. 

22.  If  two  right-angled  As  have  the  hypothenuse   and   an 

acute  angle  in  the  one  respectively  equal  to  the  like 
parts  in  the  other,  the  As  are  congruent. 


PARALLELS,   ETC.  53 

23.  The  bisectors  of  two  adjacent  angles  of  a  1      7  are  JL  to 

one  another. 

24.  ABC  is  a  A-     The  angle  between  the  external  bisector 

of  B  and  the  side  AC  is  |(C  -  A). 

25.  The  external  bisectors  of  B  and  C  meet  in  D.     Then 

^BDC  =  i(B  +  C). 

26.  A  line  L  which  coincides  with  the  side  AB  of  the  AABC 

rotates  about  B  until  it  coincides  with  BC,  without  at 
any  time  crossing  the  triangle.  Through  what  angle 
does  it  rotate  } 

27.  The  angle  required  in  Ex.  26  is  an  external  angle  of  the 

triangle.  Show  in  this  way  that  the  sum  of  the  three 
external  angles  of  a  triangle  is  a  circumangle,  and  that 
the  sum  of  the  three  internal  angles  is  a  straight  angle. 

28.  What  property  of  space  is  assumed  in  the  proof  of  Ex.  27? 

29.  Prove  76°  by  assuming  that  AC  rotates  to  AB  by  crossing 

the  triangle  in  its  rotation,  and  that  AB  rotates  to  CB, 
and  finally  CB  rotates  to  CA  in  like  manner. 


SECTION   V. 

THE  CIRCLE. 

92°.  Def.  I. — A  Circle  is  the  locus  of  a  point  which,  movin< 
in  the  plane,  keeps  at  a  constant  ____-^==-— b 

distance  from  a  fixed  point  in  the 
plane. 

The  compasses,  whatever  be  their 
form,  furnish  us  with  two  points,  A  "^^a 

and  B,  which,  from  the  rigidity  of  _ 
the  instrument,   are    supposed    to 

preserve    an    unvarying    distance  ^  ^ 

from  one  another.     Then,  if  one  of  the  points  A  is  fixed, 
while  the  other  B  moves   over  the  paper  or  other   plane 


54  SYNTHETIC   GEOMETRY. 

surface,  the  moving  point  describes  a  physical  circle.  The 
limit  of  this  physical  circle,  when  the  curved  line  has  its 
thiclcness  diminished  endlessly,  is  the  geometric  circle. 

Def.  2. — The  fixed  point  is  the  centre  of  the  circle,  and  the 
distance  between  the  fixed  and  moveable  points  is  the  radius 
of  the  circle. 

The  curve  itself,  and  especially  where  its  length  is  under 
consideration,  is  commonly  called  the  circumference  of  the 
circle. 

The  symbol  employed  for  the  circle  is  0. 

93°  From  the  definitions  of  92°  we  deduce  the  following 
corollaries  : — 

1.  All  the  radii  of  a  0  are  equal  to  one  another. 

2.  The  0  is  a  closed  figure  ;  so  that  to  pass  from  a  point 
within  the  figure  to  a  point  without  it,  or  vice  versa^  it  is 
necessary  to  cross  the  curve. 

3.  A  point  is  within  the  0,  on  the  0,  or  without  the  0, 
according  as  its  distance  from  the  centre  is  less,  equal  to,  or 
greater  than  the  radius. 

4.  Two  0s  which  have  equal  radii  are  congruent ;  for,  if 
the  centres  coincide,  the  figures  coincide  throughout  and  form 
virtually  but  one  figure. 

Def. — Circles  which  have  their  centres  coincident  are 
called  concentric  circles. 

94°.  Theore7n.—h.X\VLQ  can  cut  a  circle  in  two  points, and  in 
two  points  only. 

Proof  Since  the  0  is  a  closed  curve  (93°,  2),  a  line  which 
cuts  it  must  lie  partly  within  the  0  and  partly  without.  And 
the  generating  point  (69°)  of  the  line  must  cross  the  0  in 
passing  from  without  to  within,  and  again  in  passing  from 
within  to  without. 

.*.  a  line  cuts  a  0  at  least  twice  if  it  cuts  the  0  at  all. 


THE  CIRCLE. 


55 


Again,  since  all  radii  of  the  same  0  are  equal,  if  a  line 
could  cut  a  0  three  times,  three  equal  segments  could  be 
drawn  from  a  given  point,  the  centre  of  the  0,  to  a  given  line. 
And  this  is  impossible  (63°,  3). 

Therefore  a  line  can  cut  a  0  only  twice.  q.e.d. 

Cor.  I.  Three  points  on  the  same  circle  cannot  be  in  line  ; 
or,  a  circle  cannot  pass  through  three  points  which  are  in  line. 

95°.  Def.  I. — A  line  which  cuts  a  0  is  a  secant  or  secant-line. 

Def.  2. — The  segment  of  a  secant 
included  within  the  0  is  a  chord. 

Thus  the  line  L,  or  AB,  is  a 
secant,  and  the  segment  AB  is  a 
chord.  (21°) 

The  term  chord  whenever  involv- 
ing the  idea  of  length  means  the  segment  having  its  end- 
points  on  the  circle.  But  sometimes,  when  length  is  not 
involved,  it  is  used  to  denote  the  whole  secant  of  which  it 
properly  forms  a  part. 

Def.  3. — A  secant  which  passes  through  the  centre  is  a 
centre-line,  and  its  chord  is  a  diameter. 

Where  length  is  not  implied,  the  term  diameter  is  some- 
times used  to  denote  the  centre-Hne  of  which  it  properly 
forms  a  part. 

Thus  M  is  a  centre-line  and  CD  is  a  diameter. 


96°.     Theorejn. — Through  any  three  points  not  in  line — 

1.  One  circle  can  be  made  to  pass.  *-0 

2.  Only  one  circle  can  be  made  to  pass. 
Proof. — Let  A,  B,  C  be  three  points 

not  in  line. 

Join  AB  and  BC,and  let  L  and  M  be  the  a" 
right  bisectors  of  AB  and  BC  respectively. 


I.  Then,  because  AB  and  BC  intersect  at  B, 

L  and  M  intersect  at  some  point  O,        (79° 


Cor.) 


56  SYNTHETIC    GEOMETRY. 

and  O  is  equidistant  from  A,  B,  and  C.  (86°) 

.•.  the  0  with  centre  at  O,  and  radius  equal  to  OA,  passes 
through  B  and  C.  q.e.d. 

2.  Any  O  through  A,  B,  and  C  must  have  its  centre  equally 
distant  from  these  three  points. 

But  O  is  the  only  point  in  the  plane  equidistant  from  A,  B, 
and  C.  (86°,  Cor.) 

And  we  cannot  have  two  separate  0s  having  the  same 

centre  and  the  same  radius.  (93°,  4) 

.•.  only  one  circle  can  pass  through  A,  B,  and  C.  q.e.d. 

Cor.  I.  Circles  which  coincide  in  three  points  coincide 
altogether  and  form  one  circle. 

Cor.  2.  A  point  from  which  more  than  two  equal  segments 
can  be  drawn  to  a  circle  is  the  centre  of  that  circle. 

Cor.  3.  Since  L  is  a  centre-line  and  is  also  the  right 
bisector  of  AB, 

.•.  the  right  bisector  of  a  chord  is  a  centre  line. 

Cor.  4.  The  AAOB  is  isosceles,  since  OA=OB.  Then,  if 
D  be  the  middle  of  AB,  OD  is  a  median  to  the  base  AB  and 
is  the  right  bisector  of  AB.  (55°,  Cor.  2) 

.'.  a  centre-line  which  bisects  a  chord  is  perpendicular  to 
the  chord. 

Cor.  5.  From  Cor.  4  by  the  Rule  of  Identity, 

A  centre  line  which  is  perpendicular  to  a  chord  bisects  the 
chord. 

.*.  the  right  bisector  of  a  chord,  the  centre-line  bisecting 
the  chord,  and  the  centre-line  perpendicular  to  the  chord  are 
one  and  the  same. 

97°.  From  92°,  Def.,  a  circle  is  given  when  the  position  of 
its  centre  and  the  length  of  its  radius  are  given.  And,  from 
96°,  a  circle  is  given  when  any  three  points  on  it  are  given. 

It  will  be  seen  hereafter  that  a  circle  is  determined  by  three 
points  even  when  two  of  them  become  coincident,  and  in 
higher  geometry  it  is  shown  that  three  points  determine  a 


THE  CIRCLE.  57 

circle,  under  certain  circumstances,  when  all  three  of  the 
points  become  coincident. 

Def. — Any  number  of  points  so  situated  that  a  circle  can 
pass  through  them  are  said  to  be  concyclic^  and  a  rectilinear 
figure  (14°,  Def.)  having  its  vertices  concyclic  is  said  to  be 
inscribed  in  the  circle  which  passes  through  its  vertices,  and 
the  circle  is  said  to  circumscribe  the  figure. 

Hence  the  circle  which  passes  through  three  given  points 
is  the  circumcircle  of  the  triangle  having  these  points  as  ver- 
tices, and  the  centre  of  that  circle  is  the  circumcentre  of  the 
triangle,  and  its  radius  is  the  circumradius  of  the  triangle. 

(86°,  Def.) 

A  hke  nomenclature  applies  to  any  rectilinear  figure  having 
its  vertices  concyclic. 

98°.  Theorem. — If  two  chords  bisect  one  another  they  are 
both  diameters. 

If  AP  =  PD  and  CP  =  PB,  then  P  is  the  ''' 
centre. 

Proof. — Since  P  is  the  middle  point  of 
both  AD  and  CB  (hyp.),  therefore  the  right 
bisectors  of  AD  and  CB  both  pass  through  P. 

But  these  right  bisectors  also  pass  through  the  centre ; 
(96°,  Cor.  3)  .'.     P  is  the  centre.         (24°,  Cor.  3)  q.e.d. 

99°.   Theorem. — Equal  chords  are  equally  distant  from  the 
centre  ;   and,  conversely,  chords  equally  dis-     ^ 
tant  from  the  centre  are  equal. 
.  If  AB  =  CD  and  OE  and  OF  are  the  per- 
pendiculars   from    the    centre    upon    these 
chords,  then  OE  =  OF  ;    and  conversely,  if 
OE  =  OF,  then  AB  =  CD.  ^ 

Proof. — Since  OE  and  OF  are  centre  lines  J.  to  AB  and  CD, 
AB  and  CD  are  bisected  in  E  and  F.     (96°,  Cor.  5) 
.'.  in  the  As  OBE  and  ODF 

OB  =  OD,      EB  =  FD, 


58 


SYNTHETIC   GEOMETRY. 


and  they  are  right-angled  opposite  equal  sides, 

AOBE  =  AODF, 
and  OE  =  OF. 

Conversely,  by  the  Rule  of  Identity,  if  OE  =  OF,  then 
AB  =  CD.  q.e.d. 

ioo°.  Theorem. — Two  secants  which  make  equal  chords 
A  ^--•'" — -^B  ^p  make  equal  angles  with  the  centre-line 

through  their  point  of  intersection. 

AB  =  CD,  and  PO  is  a  centre-Hne 
through  the  point  of  intersection  of 
AB  and  CD.     Then 

^APd=ACPO. 
Proof.— L^t  OE  and   OF  be  J_  to 
AB  and  CD  from  the  centre  O. 

Then  OE  =  OF,  (99") 

AOPE  =  AOPF,        (65°) 

and  Z.APO=^CPO.  q.e.d. 

Cor.  I.   •.*  E  and  F  are  the  middle 

points  of  AB  and  CD,  (96°,  Cor.  5) 

.-.  PE  =  PF,  PA  =  PC,  and  PB  =  PD. 

Hence,  secants  which  make  equal  chords  make  two  pairs  of 

equal  line-segments  between  their  point  of  intersection  and 

the  circle. 

Cor.  2.  From  any  point  two  equal  line-segments  can  be 
drawn  to  a  circle,  and  these  make  equal  angles  with  the 
centre-line  through  the  point. 

101°.  As  all  circles  have  the  same  form,  two  circles  which 
have  equal  radii  are  equal  and  congruent  (93°,  4),  (51°). 
Hence  equal  and  cojigruent  are  equivalent  terms  when 
applied  to  the  circle. 

Def.  I, — Any  part  of  a  circle  is  an  arc. 

The  word  equal  when  applied  to  arcs  means  congruence 
or  capability  of  superposition.  Equal  arcs  come  from  the 
same  circle  or  from  equal  circles. 


THE   CIRCLE. 


59 


Def.  2.— A  line  which  divides  a  figure  into  two  parts  such 
that  when  one  part  is  revolved  about  the  line  it  may  be  made 
to  fall  on  and  coincide  with  the  other  part  is  an  axis  of 
symmetry  of  the  figure. 


102°.  Theorem. 
the  circle. 


-A  centre-line  is  an  axis  of  symmetry  of 
A^ — ^^B 


Proof. — Let  AB  and  CD  be  equal 
chords  meeting  at  P,  and  let  PHOG 
be  a  centre  line. 

Let  the  part  of  the  figure  which  lies  q> 
upon  the  F  side  of  PG  be  revolved 
about  PG  until  it  comes  to  the  plane 
on  the  E  side  of  PG. 

Then-/    ^GPA  =  ^GPC,         (100°) 
.*.  PC  coincides  with  PA. 
And       •.•  PB  =  PD 

and  PA  =  PC,  (100°,  Cor.  i) 

.*.  D  coincides  with  B, 
and  C  coincides  with  A. 

And  the  arc  HCG,  coinciding  in  three  points  with  the  arc 
HAG,  is  equal  to  it,  and  the  two  arcs  become  virtually  but 
one  arc.  (96°,  Cor.  i) 

Therefore  PG  is  an  axis  of  symmetry  of  the  O,  and  divides 
it  into  two  equal  arcs.  q.e.d. 

Def — Each  of  the  arcs  into  which  a  centre-line  divides  the 
circle  is  a  semicircle. 

Any  chord,  not  a  centre-line,  divides  the  circle  into  unequal 
arcs,  the  greater  of  which  is  called  the  major  arc,  and  the 
other  the  minor  arc. 

Cor.  I.  By  the  superposition  of  the  theorem  we  see  that 
arc  AB  =  arc  CD,  arc  HB=arc  HD,  arc  GA=arc  GC, 

arc  BDCA  =  arc  DBAC.  (ist  Fig.) 

But  the  arcs  BDCA  and  AB  are  the  major  and  minor  arcs 


60  SYNTHETIC   GEOMETRY. 

to  the  chord  AB,  and  the  arcs  DBAC  and  CD  are  major  and 
minor  arcs  to  the  chord  CD. 

Therefore  equal  chords  determine  equal  arcs,  major  being 
equal  to  major  and  minor  to  minor. 

Cor.  2.  Equal  arcs  subtend  equal  angles  at  the  centre. 

103°.  Theorem. — Parallel  secants  intercept  equal  arcs  on  a 
E  circle. 

If  AB  is  II  to  CD, 

then  arc  A C  =  arc  D  B. 


/d      Proof. — Let  EF  be  the  centre-line  _L  to  AB. 
Then  EF  is  J_  to  CD  also.  (72^) 

When  EBDF  is  revolved  about  EF, 
B  comes  to  coincidence  with  A,  and  D  with  C,  and  the  arc 
BD  with  the  arc  AC, 

arc  AC  =  arc  DB.  q.e.d. 

Cor.  Since  the  chord  AC  =  chord  BD, 
Therefore  parallel  chords  have  the   chords  joining  their 
end-points  equal. 

Exercises. 

1.  Any  plane  closed  figure  is  cut  an  even  number  of  times  by 

an  indefinite  line. 

2.  In  the  figure  of  Art.  96°,  if  A,  B,  and  C  shift  their  relative 

positions  so  as  to  tend  to  come  into  line,  what  becomes 
of  the  point  O  ? 

3.  In  the  same  figure,  if  ABC  is  a  right  angle  where  is  the 

point  O  } 

4.  Given  a  circle  or  a  part  of  a  circle,  show  how  to  find  its 

centre. 

5.  Three  equal  segments  cannot  be  drawn  to  a  circle  from 

a  point  without  it. 

6.  The  vertices  of  a  rectangle  are  concyclic. 

7.  If  equal  chords  intersect,  the  segments  of  one  between  the 


THE   CIRCLE.  6 1 

point  of  intersection  and  the  circle  are  respectively 
equal  to  the  corresponding  segments  of  the  other. 

8.  Two  equal  chords  which  have  one  end-point  in  common 

lie  upon  opposite  sides  of  the  centre. 

9.  If  AB  and  CD  be  parallel  chords,  AD  and  BC,  as  also 

AC  and  BD,  meet  upon  the  right  bisector  of  AB  or  CD. 

10.  Two  secants  which  make  equal  angles  with  a  centre-line 

make  equal  chords  in  the  circle  if  they  cut  the  circle. 

(Converse  of  100°) 

11.  What  is  the  axis  of  symmetry  of  (a)  a  square,   (d)  a 

rectangle,  (c)  an  isosceles  triangle,  (d)  an  equilateral 
triangle  ?  Give  all  the  axes  where  there  are  more  than 
one. 

12.  When  a  rectilinear  figure  has  more  than  one  axis  of 

symmetry,  what  relation  in  direction  do  they  hold  to 
one  another  ? 

13.  The  vertices  of  an  equilateral  triangle  trisect  its  circum- 

circle. 

14.  A  centre-line  perpendicular  to  a  chord  bisects  the  arcs 

determined  by  the  chord. 

1 5.  Show  how  to  divide  a  circle  (a)  into  6  equal  parts,  {d)  into 

8  equal  parts. 

16.  If  equal  chords  be  in  a  circle,  one  pair  of  the  connecters 

of  their  end-points  are  parallel  chords. 

(Converse  of  103°,  Cor.) 


THE  PRINCIPLE  OF  CONTINUITY. 

104°.  The  principle  of  continuity  is  one  of  the  most  prolific 
in  the  whole  range  of  Mathematics. 

Illustrations  of  its  meaning  and  application  in  Geometr}^ 
will  occur  frequently  in  the  sequel,  but  the  following  are 
given  by  way  of  introduction. 

I.  A  magnitude  is  cojitinuous  throughout  its  extent. 

Thus  a  line  extends  from  any  one  point  to  another  without 


62  SYNTHETIC  GEOMETRY. 

breaks  or  interruptions  ;  or,  a  generating  point  in  passing 
from  one  position  to  another  must  pass  through  every  inter- 
mediate position. 

2.  In  Art.  53°  we  have  the  theorem — Every  point  on  the 
right  bisector  of  a  segment  is  equidistant  from  the  end-points 
of  the  segment. 

In  this  theorem  the  Imiiting  condition  in  the  hypothesis  is 
that  the  point  must  be  on  the  right  bisector  of  the  segment. 

Now,  if  P  be  any  point  on  the  right  bisector,  and  we  move 
P  along  the  right  bisector,  the  limiting  condition  is  not  at 
any  time  violated  during  this  motion,  so  that  P  remains  con- 
tijtuously  equidistant  from  the  end-points  of  the  segment 
during  its  motion. 

We  say  then  that  the  property  expressed  in  the  theorem  is 
continuous  while  P  moves  along  the  right  bisector. 

3.  In  Art.  97°  we  have  the  theorem — The  sum  of  the  in- 
ternal angles  of  a  quadrangle  is  four  right  angles. 

The  limiting  condition  is  that  the  figure  shall  be  a  quad- 
A  rangle,  and  that  it  shall  have  in- 

ternal angles. 

Now,  let  ABCD  be  a  quadrangle. 
Then  the  condition  is  not  violated 
if  D  moves  to  D^  or  D2.     But  in 
^  the  latter  case  the  normal  quad- 

rangle ABCD  becomes  the  inverted  quadrangle  ABCDg, 
and  the  theorem  remains  true.  Or,  the  theorem  is  continu- 
ously true  while  the  vertex  D  moves  anywhere  in  the  plane, 
so  long  as  the  figure  remains  a  quadrangle  and  retains  four 
internal  angles. 

Future  considerations  in  which  a  wider  meaning  is  given 
to  the  word  "  angle  "  will  show  that  the  theorem  is  still  true 
even  when  D,  in  its  motion,  crosses  one  of  the  sides  AB  or 
BC,  and  thus  produces  the  crossed  quadrangle. 

The  Principle  of  Continuity  avoids  the  necessity  of  proving 
theorems  for  different  cases  brought  about  by  variations  in 
the  disposition  of  the  parts  of  a  diagram,  and  it  thus  gener- 


THE  CIRCLE.  63 

alizes  theorems  or  relieves  them  from  dependence  upon  the 
particularities  of  a  diagram.  Thus  the  two  figures  of  Art. 
100°  differ  in  that  in  the  first  figure  the  secants  intersect 
without  the  circle,  and  in  the  second  figure  they  intersect 
within,  while  the  theorem  applies  with  equal  generality  to  both. 
The  Principle  of  Continuity  may  be  stated  as  follows  : — 
When  a  figure,  which  involves  or  illustrates  some  geometric 
property,  can  undergo  change,  however  small,  in  any  of  its 
parts  or  in  their  relations  without  violating  the  conditions 
upon  which  the  property  depends,  then  the  property  is  con- 
tinuous while  the  figure  undergoes  any  amount  of  change  of 
the  same  kind  within  the  range  of  possibility. 

105°.  Let  AB  be  a  chord  dividing  the  0  into  unequal  arcs, 
and  let  P  and  Q  be  any  points  upon 
the  major  and  minor  arcs  respectively. 
(102°,  Def ) 

Let  O  be  the  centre. 

1.  The  radii  OA  and  OB  form  two 
angles  at  the  centre,  a  major  angle 
denoted  by  a  and  a  minor  angle  de- 
noted by  ^.  These  together  make  up 
a  circumangle. 

2.  The  chords  PA,  PB,  and  QA,  QB  form  two  angles  at 
the  circle,  of  which  APB  is  the  tninor  angle  and  AQB  is  the 
major  angle. 

3.  The  minor  angle  at  the  circle,  APB,  and  the  minor  angle 
at  the  centre,  /3,  stand  upon  the  minor  arc,  AQB,  as  a  base. 
Similarly  the  major  angles  stand  upon  the  major  arc  as  base. 

4.  Moreover  the  Z-APB  is  said  to  be  z;z  the  arc  APB,  so 
that  the  minor  angle  at  the  circle  is  in  the  major  arc,  and  the 
major  angle  at  the  circle  is  in  the  minor  arc. 

5.  When  B  moves  towards  B'  all  the  minor  elements 
increase  and  all  the  major  elements  decrease,  and  when  B 
comes  to  B'  the  minor  elements  become  respectively  equal  to 
the  major,  and  there  is  neither  major  nor  minor. 


64  SYNTHETIC   GEOMETRY. 

When  B,  moving  in  the  same  direction,  passes  B',  the 
elements  change  name,  those  which  were  formerly  the  minor 
becoming  the  major  and  vice  versa. 

io6°.  Theorem. — An  angle  at  the  circle  is  one-half  the 
corresponding  angle  at  the  centre,  major  corresponding  to 
major  and  minor  to  minor. 

p  Z.AOB  minor  is  2Z.APB. 

/^     /\  \    N^  Proof. — Since  AAPO  is  isosceles, 

/      / a'l    \'Y    .'.  /.OAP  =  z.OPA,(53°,Cor.i) 

/j^/sC\         and^0AP  +  ^0PA=2^0PA. 
}^:^^T^y       ^"^  zj\OC=^OAP  +  ^OPA, 

\>1      y/^  (76°,  Cor.) 

^^~p^  .*.  ^A0C  =  2^0PA. 

Similarly  ^B0C  =  2^0PB  ; 

.-.adding,  ^AOB  minor  =  2^APB.  g.e.d. 

The  theorem  is  thus  proved  for  the  minor  angles.  But 
since  the  limiting  conditions  require  only  an  angle  at  the 
circle  and  an  angle  at  the  centre,  the  theorem  remains  true 
while  B  moves  along  the  circle.  And  when  B  passes  B' 
the  angle  APB  becomes  the  major  angle  at  the  circle,  and 
the  angle  AOB  minor  becomes  the  major  angle  at  the 
centre. 

the  theorem  is  true  for  the  major  angles. 

Cor.  I.  The  angle  in  a  given  arc  is  constant.  (105°,  4) 

Cor.  2.  Since      ^APB  =  |z.AOB  minor, 
and  ^AOB  =  i^OB  major, 

and  '.•  ^AOB  minor +  Z.AOB  major  =  4  right  angles        (37°) 
Z.APB  +  ^QB  =  a  straight  angle. 

And  APBQ  is  a  coney clic  quadrangle. 
Hence   a   concyclic   quadrangle  has    its    opposite   internal 
angles  supplementary.  (40°,  Def.  i) 

Cor.  3.  D  being  on  AQ  produced, 

lBQD  is  supplementary  to  ^AOB. 


THE   CIRCLE. 


65 


But      Z.APB  is  supplementary  to  ^A(2B, 
^APB=Z-BOD. 
Hence,  if  one  side  of  a  concyclic  quadrangle 
be  produced,  the  external  angle  is  equal  to  a 
the  opposite  internal  angle. 

Cor.  4.  Let  B  come  to  B'.         (Fig.  of  106°) 
Then  ^AOB'  is  a  straight  angle, 

Z.APB' is  a  right  angle. 
But  the  arc  APB'  is  a  semicircle, 


(102°,  Def ) 


Therefore  the  angle  in  a  semicircle  is  a  right  angle. 


opposite 


107°.   Theorem. — A   quadrangle    which    has    its 
angles  supplementary  has  its  vertices  concyclic. 

(Converse  of  106° 

ABCD  is  a  quadrangle  whereof  the  ^lADC 
is  supplementary  to  ^ABC  ;  then  a  circle 
can  pass  through  A,  B,  C,  and  D. 

Proof. — If  possible  let  the  0  through  A, 
B,  and  C  cut  AD  in  some  point  P. 

Join  P  and  C. 

Then  Z_APC  is  supplementary  to  Z.ABC,  (106°,  Cor.  2) 
and  zADC  is  supplementary  to  Z.ABC, 

z.APC=^DC, 
which  is  not  true. 

.*.  the  0  cannot  cut  AD  in  any  point  other  than  D, 
Hence  A,  B,  C,  and  D  are  concyclic. 

Cor.  I.  The  hypothenuse  of  a  right-angled  triangle  is  the 
diameter  of  its  circumcircle.  (88°,  3,  Def. ;  97°,  Def.) 

Cor.  2.  When  P  moves  along  the  0  the  AAPC  (last  figure) 
has  its  base  AC  constant  and  its  vertical  angle  APC  constant. 

Therefore  the  locus  of  the  vertex  of  a  triangle  which  has  a 
constant  base  and  a  constant  vertical  angle  is  an  arc  of  a 
circle  passing  through  the  end-points  of  the  base. 

This  property  is  employed  in  the  trammel  which  is  used  to 
describe  an  arc  of  a  given  circle. 

E 


(hyp.) 
(60^) 
g.e.d. 


66 


SYNTHETIC   GEOMETRY. 


It  consists   of  two  rules   (i6°)    L    and    M    joined  at  a 

L  ^y^2\  determined  angle.     When  it  is  made 

to   slide   over  two  pins  A  and   B,  a 

pencil    at   P   traces    an    arc    passing 

through  A  and  B. 

io8°.  Theorem. — The  angle  between  two  intersecting  se- 
cants is  the  sum  of  those  angles  in  the  circle  which  stand 
on  the  arcs  intercepted  between  the 
secants,  when  the  secants  intersect 
within  the  circle,  and  is  the  difference 
of  these  angles  when  the  secants 
intersect  without  the  circle. 

lAPC=^ABC  +  z_BCD,  {i%) 
^APC  =  aABC-^BCD.  CfIJ) 
Proof.-i.  z.APC=^PBC+^PCB,  (6o°) 
.-.      aAPC  =  z.ABC  +  £.BCD. 
2.  /.ABC=^APC  +  ^BCP, 
.-.      ^APC  =  ^ABC-^BCD. 

q.e.d. 


Exercises. 

1.  If  a  six-sided  rectilinear  figure  has  its  vertices  concyclic, 

the  three  alternate  internal  angles  are  together  equal 
to  a  circumangle. 

2.  In  Fig.  105°,  when  B  comes  to  O,  BQ  vanishes ;  what  is 

the  direction  of  BQ  just  as  it  vanishes  1 

3.  Two  chords  at  right  angles  determine  four  arcs  of  which  a 

pair  of  opposite  ones  are  together  equal  to  a  semi- 
circle. 

4.  A,  B,  C,  D  are  the  vertices  of  a  square,  and  A,  E,  F  of  an 

equilateral  triangle  inscribed  in  the  same  circle. 
What  is  thQ  angle  between  the  lines  BE  and  DF? 
between  BF  and  ED  ? 


THE  CIRCLE. 


67 


SPECIAL  SECANTS— TANGENT. 


Let  P  be  a  fixed  point  on  the  0  S  and  O  a  variable 


109 
one. 

The  position  of  the  secant  L,  cut- 
ting the  circle  in  P  and  Q,  depends 
upon  the  position  of  O. 

As  Q  moves  along  the  0  the  secant 
rotates  about  P  as  pole.  While  Q 
makes  one  complete  revolution  along 
the  0  the  secant  L  passes  through  two 
special  positions.  The  first  of  these  is  when  O  is  farthest 
distant  from  P,  as  at  Q',  and  the  secant  L  becomes  a  centre- 
line. The  second  is  when  Q  comes  into  coincidence  with  P, 
and  the  secant  takes  the  position  TT'  and  becomes  a  tangent. 

Def.  I.— A  tangent  to  a  circle  is  a  secant  in  its  limiting 
position  when  its  points  of  intersection  with  the  circle  become 
coincident. 

That  the  tangent  cannot  cut  or  cross  the  0  is  evident. 
For  if  it  cuts  the  0  at  P  it  must  cut  it  again  at  some  other 
point.  And  since  P  represents  two  points  we  would  have  the 
absurdity  of  a  line  cutting  a  circle  in  three  points.  (94°) 

Def.  2.— The  point  where  P  and  Q  meet  is  called  Xh^  point 
of  contact.  Being  formed  by  the  union  of  two  points  it  repre- 
sents both,  and  is  therefore  a  double  point. 

From  Defs.  i  and  2  we  conclude — 

1.  A  point  of  contact  is  a  double  point. 

2.  As  a  line  can  cut  a  0  only  twice  it  can  touch  a  0  only  once. 

3.  A  line  which  touches  a  0  cannot  cut  it. 

4.  A  0  is  determined  by  two  points  if  one  of  them  is  a 

given  point  of  contact  on  a  given  line  ;  or,  only  one 
circle  can  pass  through  a  given  point  and  touch  a 
given  line  at  a  given  point.     (Compare  97°.) 


6S 


SYNTHETIC   GEOMETRY 


iio°.   Theorem. — A  centre-line  and  a  tangent  to  the  same 
point  on  a  circle  are  perpendicular  to  one  another. 

L'  is  a  centre-line   and  T  a  tan- 
iQ  gent,  both  to  the  point  P.     Then  L' 

is  _L  to  T. 

Proof. — '.*  T  has  only  the  one  point 
P  in  common  with  the  0,  every  point 
of  T  except  P  lies  without  the  0.     .'. 
T'  if  O  is  the  centre  on  the  line  L',  OP 
is  the  shortest  segment  from  O  to  T. 
OP,  or  L',  is  ±  to  T.  (63°,  i)  q.c.cL 

Cor.  I.  Tangents   at   the  end-points  of   a    diameter  are 
parallel. 

Cor.  2.  The  perpendicular  to  a  tangent  at  the  point  of  con- 
tact is  a  centre-line.     (Converse  of  the  theorem.) 

Cor.  3.  The  perpendicular  to  a  diameter  at  its  end-point  is 
a  tangent. 


111°.  Theorem. — The   angles  between  a    tangent  and  a 

D  chord  from  the   point    of   contact   are 

respectively  equal  to  the  angles  in  the 

opposite    arcs    into    which    the    chord 

divides  the  circle. 

TP  is  a  tangent  and  PQ  a  chord  to 
the  same  point  P,  and  A  is  any  point  on 
the  0.     Then 

^QPT=^QAP. 

Let  PD  be  a  diameter. 
^QAP  =  ^QDP, 
/.DQP  is  a  "X 
^DPQ  is  comp.  of  Z.QPT, 
^DPQ  is  comp.  of  ^ODP, 
^QDP=i.QPT=z.QAP. 
Similarly,  the    ^QPT'=^QBP. 


Proof.- 
Then 
and 
Also 
and 


(106°,  Cor.  i) 

(106°,  Cor.  4) 

(40°,  Def.  3) 

{n%  6) 

q.e.d. 


THE  CIRCLE. 


69 


112°.  Theorem. — Two  circles  can  intersect  in  only  two 
points. 

Proof. — If  they  can  intersect  in  three  points,  two  circles 
can  be  made  to  pass  through  the  same  three  points.  But 
this  is  not  true.  (96°) 

.'.  two  circles  can  intersect  in  only  two  points. 

Cor.  Two  circles  can  touch  in  only  one  point.  For  a 
point  of  contact  is  equivalent  to  two  points  of  intersection. 


113°.  Theorem. — The  common  centre-line  of  two  intersect- 
ing circles  is  the  right  bisector 
of  their  common  chord. 

O  and  O'  are  the  centres  of  S 
and  S',  and  AB  is  their  common 
chord.  Then  00'  is  the  right 
bisector  of  AB. 


(50 


Proof.-  Since  AO  =  BO, 

and  AO'  =  BO', 

.*.     O  is  on  the  right  bisector  of  AB. 
Similarly  O'  is  on  the  right  bisector  of  AB, 

.*.     00'  is  the  right  bisector  of  AB. 

Cor.  I.  By  the  principle  of  continuity,  OO'  always  bisects 
AB.  Let  the  circles  separate  until  A  and  B  coincide.  Then 
the  circles  touch  and  00'  passes  through  the  point  of  contact. 

Def. — Two  circles  which  touch  one  another  have  external 
contact  when  each  circle  lies  without  the  other,  and  internal 
contact  when  one  circle  lies  within  the  other. 

Cor.  2.  Since  00'  (Cor.  i)  passes  through  the  point  of 
contact  when  the  circles  touch  one  another — 

{(.i)  When  the  distance  between  the  centres  of  two  circles 

is  the  sum  of  their  radii,  the  circles  have  external 

contact. 
{b)  When  the  distance  between  the  centres  is  the  difference 

of  the  radii,  the  circles  have  internal  contact. 


70 


SYNTHETIC   GEOMETRY. 


(c)  When  the  distance  between  the  centres  is  greater  than 

the  sum  of  the  radii,  the  circles  exclude  each  other 
without  contact. 

(d)  When  the  distance  between  the  centres  is  less  than 

the  difference  of  the  radii,  the  greater  circle  includes 
the  smaller  without  contact. 

(e)  When  the  distance  between  the  centres  is  less  than  the 

sum  of  the  radii  and  greater  than  their  difference,  the 
circles  intersect. 


14°.  Theo7-em. — From  any  point  without  a  circle  two 
tangents  can  be  drawn  to  the 
circle. 

Proof.— 1.^1  S  be  the  0  and  P 
the  point.  Upon  the  segment 
PO  as  diameter  let  the  ©S'  be 
described,  cutting  0S  in  A  and 
B.  Then  PA  and  PB  are  both 
tangents  to  S. 

For  Z.OAP  is  in  a  semicircle  and  is  a  "~|'  (106°,  Cor.  4) 

.-.  AP  is  tangent  to  S.  (110°,  Cor.  3) 

Similarly  BP  is  tangent  to  S. 

Cor.  I.  Since  PO  is  the  right  bisector  of  AB,  (113°) 

PA  =  PB.  (53°) 

Hence  calling  the  segment  PA  the  tangent  from  P  to  the 
circle,  when  length  is  under  consideration,  we  have — The  two 
tangents  from  any  point  to  a  circle  are  equal  to  one  another. 

Def — The  line  AB,  which  passes  through  the  points  of 
contact  of  tangents  from  P,  is  called  the  chord  of  contact  for 
the  point  P. 


115°.  Def  I. — The  angle  at  which  two  circles  intersect  is 
the  angle  between  their  tangents  at  the  point  of  intersection. 

Def  2.  When  two  circles  intersect  at  right  angles  they  are 
said  to  cut  each  other  orthogonally. 


THE   CIRCLE. 


71 


The  same  term  is  conveniently  applied  to  the  intersection 
of  any  two  figures  at  right  angles. 

Cor.  I.  If,  in  the  Fig.  to  114°,  PA  be  made  the  radius  of  a 
circle  and  P  its  centre,  the  circle  will  cut  the  circle  S  ortho- 
gonally. For  the  tangents  at  A  are  respectively  perpendicular 
to  the  radii. 

Hence  a  circle  S  is  cut  orthogonally  by  any  circle  having 
its  centre  at  a  point  without  S  and  its  radius  the  tangent 
from  the  point  to  the  circle  S . 


116°.  The  following  examples  furnish  theorems  of  some 
importance. 

Ex.  I.  Three  tangents  touch  the  circle 
S  at  the  points  A,  B,  and  C,  and  inter- 
sect to  form  the  AA'B'C.  O  being  the 
centre  of  the  circle, 

A.A0C  =  2^A'0C'. 

Proof.—         AC'=BC', 
and  BA'  =  CA',  (114°,  Cor.  i) 

AAOC'  =  ABOC,  and  ABOA'  =  ACOA' 
^BOC'  =  ^AOC,   and   ^COA'=^BOA', 
zAOC  =  2^A'OC'.  q.e.d. 

Similarly        ^AOB  =  2/.A'OB',  and  ^B0C  =  2^B'0C'. 

If  the  tangents  at  A  and  C  are  fixed,  and  the  tangent  at  B 
is  variable,  we  have  the  following  theorem  : — 

The  segment  of  a  variable  tangent  intercepted  by  two  fixed 
tangents,  all  to  the  same  circle,  subtends  a  fixed  angle  at  the 
centre. 

Ex.  2.  If  four  circles  touch  two  and  two  externally,  the 
points  of  contact  are  concyclic. 

Let  A,  B,  C,  D  be  the  centres  of  the  circles,  and  P,  Q,  R, 
S  be  the  points  of  contact. 

Then  AB  passes  through  P,  BC  through  Q,  etc.  (113°,  Cor.  i ) 


J 2  SYNTHETIC   GEOMETRY. 

Now,  ABCD  being  a  quadrangle, 

^A  +  i.B  +  ^C  +  ^D=4~|s.  (90°) 

But  the  sum  of  all  the  internal  angles  of  the  four  As  APS, 
BQP,  CRQ,  and  DSR  is  8~ls,  and 

z.APS=^ASP,  ^BPQ=/.BQP,  etc., 

z.APS  +  ^BPQ  +  ^CRQ  +  /.DRS  =  2~ls. 
Now  *       /.SPQ  =  2~]s-(^APS  +  £.BPQ), 

^QRS  =  2  "Is  -  (^CRQ  +  Z.DRS), 
^SP0  +  /.QRS  =  2"ls. 
and  P,  Q,  R,  S  are  concyclic  (107°).  q.e.d. 

Ex.  3.  If  the  common  chord  of  two  intersecting  circles 
subtends  equal  angles  at  the  two  circles,  the  circles  are  equal. 

AB  is  the  common  chord,  C,  C  points  upon  the  circles,  and 
^ACB  =  .lACB. 

Let  O,  O'  be  the  centres.  Then  ^AOB  =  ^AO'B.  (106°) 
And  the  triangles  OAB  and  O'AB  being  isosceles  are  con- 
gruent,        .'.  OA  =  O^A,  and  the  circles  are  equal.       (93°,  4) 

Ex.  4.  If  O  be  the  orthocentre  of  a  AABC,  the  circum- 
circles  to  the  As  ABC,  AOB,  BOC,  COA 
are  all  equal. 

*.'  AX  and  CZ  are  ±  respectively  to 
BC  and  AB, 

£.CBA  =  sup.  of  ^XOZ 
=  sup.  of  ^COA. 
But   D    being    any  point  on  the    arc 
AS2C,  ^CDA  is  the  sup.  of  ^COA. 

^CBA  =  ^CDA, 
and  the  0s  S  and  83  are  equal  by  Ex.  3. 
In  like  manner  it  may  be  proved  that  the  0S  is  equal  to  the 
03  S3  and  Si. 

Ex.  5.  If  any  point  O  be  joined  to  the  vertices  of  a  AABC, 
the  circles  having  OA,  OB,  and  OC  as  diameters  intersect 
upon  the  triangle. 

Proof.— VixTs.^  OX  _L  to  BC  and  OY  ±  to  AC. 


THE   CIRCLE. 


73 


•/  z.OXB=~~|,  the  0  on  OB  as  diameter  passes  through  X. 

(107°,  Cor.  I) 
Similarly  the  0  on  OC  as  diameter  passes  through  X. 
Therefore  the  0s  on  OB  and  OC  intersect  in  X  ;  and  in  like 
manner  it  is  seen  that  the  0s  on  OC  and  OA  intersect  in  Y, 
and  those  on  OA  and  OB  intersect  in  Z,  the  foot  of  the  X 
from  O  to  AB. 

Ex.  6.  The  feet  of  the  medians  and  the  feet  of  the  altitudes 
in  any  triangle  are  six  concyclic  points,  and  the  circle  bisects 
that  part  of  each  altitude  lying 
between  the  orthocentre  and  the 
vertex. 

D,  E,  F  are  the  feet  of  the 
medians,  i.e.,  the  middle  points 
of  the  sides  of  the  AABC.  Let 
the  circle  through  D,  E,  F  cut 
the  sides  in  G,  H,  K. 

Now  FD  is  II  to  AC  and  ED  is  ||  to  AB, 
^FDE  =  ^FAE. 
But  ^FDE  =  ^FHE, 

.-.  AAFH  is  isosceles,  and  AF  =  FH  =  FB 

^AHB=n, 
and  H  is  the  foot  of  the  altitude  from  B. 

Similarly,  K  and  G  are  feet  of  the  altitudes  from  C  and  A. 

Again,  ^KPH  =  ^KFH  =  2/.KAH.  And  A,  K,  O,  H  are 
concyclic  (107°),  and  AO  is  a  diameter  of  the  circumcircle, 
therefore  P  is  the  middle  point  of  AO. 

Similarly,  O  is  the  middle  point  of  BO,  and  R  of  CO. 

Def. — The  circle  S  passing  through  the  nine  points  D,  E, 
F,  G,  H,  K,  and  P,  Q,  R,  is  called  the  7iine-poiiits  circle  of 
the  AABC. 

Cor.  Since  the  nine-points  circle  of  ABC  is  the  circum- 
circle of  ADEF,  whereof  the  sides  are  respectively  equal  to 
half  the  sides  of  the  AABC,  therefore  the  radius  of  the  nine- 
points  circle  of  any  triangle  is  one-half  that  of  its  circumcircle. 


(84°,  Cor.  2) 
(106°,  Cor.  I) 
(106°,  Cor.  4) 


74  SYNTHETIC  GEOMETRY. 


Exercises. 


1.  In  105°  when  P  passes  B  where  is  the  Z.APB  ? 

2.  A,  B,  C,  D  are  four  points  on  a  circle  whereof  CD  is  a 

diameter  and  E  is  a  point  on  this  diameter.  If 
z_AEB  =  2^ACB,  E  is  the  centre. 

3.  The  sum  of  the  alternate  angles  of  any  octagon  in  a  circle 

is  six  right  angles. 

4.  The  sum  of  the  alternate  angles  of  any  concyclic  polygon 

of  2;/  sides  is  2{n  —  i)  right  angles. 

5.  If  the  angle  of  a  trammel  is  60°  what  arc  of  a  circle  will 

it  describe  ?  what  if  its  angle  is  7t°  ? 

6.  Trisect  a  right  angle  and  thence  show  how  to  draw  a 

regular  12-sided  polygon  in  a  circle. 

7.  If  r,  r'  be  the  radii  of  two  circles,  and  d  the  distance 

between  them,  the  circles  touch  when  d=r±r'. 

8.  Give  the  conditions  under  which  two  circles  have  4,  3,  2, 

or  I  common  tangent. 

9.  Prove  Ex.  2,  116°,  by  drawing  common  tangents  to  the 

circles  at  P,  Q,  R,  and  S. 
I  o.  A  variable  chord  passes  through  a  fixed  point  on  a  circle, 
to  find  the  locus  of  the  middle  point  of  the  chord. 

11.  A  variable  secant  passes  through  a  fixed  point,  to  find 

the  locus  of  the  middle  point  of  the  chord  determined 
by  a  fixed  circle. 

12.  In  Ex.  II,  what  is  the  locus  of  the  middle  point  of  the 

secant  between  the  fixed  point  and  the  circle  ? 

13.  In  a  quadrangle  circumscribed  to  a  circle  the  sums  of  the 

opposite  sides  are  equal  in  pairs  ;  and  if  the  vertices 
be  joined  to  the  centre  the  sums  of  the  opposite  angles 
at  the  centre  are  equal  in  pairs. 

14.  If  a  hexagon  circumscribe  a  circle  the  sum   of  three 

alternate  sides  is  equal  to  that  of  the  remaining 
three. 

15.  If  two  circles  are  concentric,  any  chord  of  the  outer 

which  is  tangent  to  the  inner  is  bisected  by  the  point 


THE   CIRCLE.  75 

of  contact ;  and  the  parts  intercepted  on  any  secant 
between  the  two  circles  are  equal  to  one  another. 

1 6.  If  two  circles  touch  one  another,  any  line  through  the 

point  of  contact  determines  arcs  which  subtend  equal 
angles  in  the  two  circles. 

17.  If  any  two  lines  be  drawn  through  the  point  of  contact  of 

two  touching  circles,  the  lines  determine  arcs  whose 
chords  are  parallel. 

18.  If  two  diameters  of  two  touching  circles  are  parallel,  the 

transverse  connectors  of  their  end-points  pass  through 
the  point  of  contact. 

19.  The  shortest  chord  that  can  be  drawn  through  a  given 

point  within  a  circle  is  perpendicular  to  the  centre-line 
through  that  point. 

20.  Three  circles  touch  each  other  externally  at  A,  B,  and  C. 

The  chords  AB  and  AC  of  two  of  the  circles  meet  the 
third  circle  in  D  and  E.  Prove  that  DE  is  a  diameter 
of  the  third  circle  and  parallel  to  the  common  centre- 
line of  the  other  two. 

21.  A  line  which  makes  equal  angles  with  one  pair  of  oppo- 

site sides  of  a  concyclic  quadrangle  makes  equal  angles 
with  the  other  pair,  and  also  with  the  diagonals. 

22.  Two  circles  touch  one  another  in  A  and  have  a  common 

tangent  BC.     Then  lBAC  is  a  right  angle. 

23.  OA  and  OB  are  perpendicular  to  one  another,  and  AB  is 

variable  in  position  but  of  constant  length.  Find  the 
locus  of  the  middle  point  of  AB. 

24.  Two  equal  circles  touch  one  another  and  each  touches 

one  of  a  pair  of  perpendicular  lines.  What  is  the  locus 
of  the  point  of  contact  of  the  circles  ? 

25.  Two  lines  through  the  common  points  of  two  intersecting 

circles  determine  on  the  circles  arcs  whose  chords  are 
parallel. 

26.  Two  circles  intersect  in  A  and  B,  and  through  B  a  secant 

cuts  the  circles  in  C  and  D.  Show  that  Z.CAD  is 
constant,  the  direction  of  the  secant  being  variable. 


jG  SYNTHETIC   GEOMETRY. 

27.  At  any  point  in  the  circumcircle  of  a  square  one  of  the 

sides  subtends  an  angle  three  times  as  great  as  that 
subtended  by  the  opposite  side. 

28.  The  three  medians  of  any  triangle  taken  in  both  length 

and  direction  can  form  a  triangle. 


SECTION   VI. 
CONSTRUCTIVE   GEOMETRY, 

INVOLVING    THE    PRINCIPLES    OF    THE    FIRST    FIVE 
SECTIONS,    ETC. 

117°.  Constructive  Geometry  applies  to  the  determination 
of  geometric  elements  which  shall  have  specified  relations  to 
given  elements. 

Constructive  Geometry  is  Practical  when  the  determined 
elements  are  physical,  and  it  is  Theoretic  when  the  elements 
are  supposed  to  be  taken  at  their  limits,  and  to  be  geometric 
in  character.  (12°) 

Practical  Constructive  Geometry,  or  simply  "  Practical 
(Geometry,"  is  largely  used  by  mechanics,  draughtsmen,  sur- 
veyors, engineers,  etc.,  and  to  assist  them  in  their  work 
numerous  aids  known  as  "  Mathematical  Instruments  "  have 
been  devised. 

A  number  of  these  will  be  referred  to  in  the  sequel. 

In  "Practical  Geometry "  the  "  Rule "  (16°)  furnishes  the 
means  of  constructing  a  line,  and  the  "  Compasses  "  (92°)  of 
constructing  a  circle. 

In  Theoretic  Constructive  Geometry  we  assume  the  ability 
to  construct  these  two  elements,  and  by  means  of  these  we 
are  to  determine  the  required  elements. 

118'.     To  test  the  "  Rule." 

Place  the  rule  on  a  plane,  as  at  R,  and  draw  a  line  AB 


CONSTRUCTIVE   GEOxMETRY.  7/ 

along  its  edge.     Turn  the  rule  into  the  position  R'.     If  the 

edge  now  coincides  with  the  line       ^ -; v^_^ 

the  rule  is  true.  ^^(7 r -^^ — 

This  test  depends  upon  the  pro- 
perty that  two  finite  points  A  and  B  determine  one  line. 

(24°,  Cor.  2) 

Def. — A  construction  proposed  is  in  general  called  a 
proposition  (2°)  and  in  particular  7ip7'oblem. 

A  complete  problem  consists  of  (i)  the  statement  of  what 
is  to  be  done,  (2)  the  construction,  and  (3)  the  proof  that  the 
construction  furnishes  the  elements  sought. 

119°.  Problem. — To  construct  the  right  bisector  of  a  given 
line  segment. 

Let  AB  be  the  given  segment. 

Construction. — With  A  and  B  as  centres 
and  with  a  radius  AD  greater  than  half  of  a" 
AB  describe  circles. 

Since  AB  is  <  the  sum  of  the  radii  and 
>  their  difference,  the  circles  will  meet  in 
two  points  P  and  Q.  (ii3°>  Cor.  2,  e) 

The  line  PQ  is  the  right  bisector  required. 

Proof. — P  and  O  are  each  equidistant  from  A  and  B  and 
.'.  they  are  on  the  right  bisector  of  AB  ;  (54°) 

.*.  PQ  is  the  right  bisector  of  AB. 

Cor.  I.  The  same  construction  determines  C,  the  middle 
point  of  AB. 

Cor.  2.  If  C  be  a  given  point  on  a  line,  and  we  take  A  and 
B  on  the  line  so  that  CA=CB,  then  the  right  bisector  of  the 
segment  AB  passes  through  C  and  is  J_  to  the  given  line. 

.•.  the  construction  gives  the  perpendicular  to  a  given  line 
at  a  given  point  in  the  line. 

120°.  Problem. — To  draw  a  perpendicular  to  a  given  line 
from  a  point  not  on  the  line. 


I  \ 
I  \ 
I       > 

^ — Id 


7S 


SYNTHETIC   GEOMETRY. 


Let  L  be  the  given  line  and  P  be  the  point. 

Constr. — Draw  any  line  through 
P  meeting  L  at  some  point  A. 
Bisect  AP  in  C  (119°,  Cor.  i),  and 
with  C  as  centre  and  CP  as  radius 
describe  a  circle. 

If  PA  is  not  i.  to  L,  the  0  will 
-cut  L  in  two  points  A  and  D. 
Then  PD  is  the  J.  required. 
Proof. — PDA  is  the  angle  in  a  semicircle, 

^PDA  is  a  "~I-  (106°,  Cor.  4) 

Cor.  Let  D  be  a  given  point  in  L.     With  any  centre  C 

and  CD  as  radius  describe  a  circle  cutting  L  again  in  some 

point  A.     Draw  the  radius  ACP,  and  join  D  and  P.     Then 

DP  is  ±  to  L. 

.*.  the  construction  draws  a  X  to  L  at  a  given  point  in  L. 

(Compare  119°,  Cor.  2) 

Cor.  2.  Let  L  be  a  given  line  and  C  a  given  point. 

To  draw  through  C  a  line  parallel  to  L. 

With  C  as  the  centre  of  a  circle,  construct  a  figure  as 

given.     Bisect  PD  in  E  (119°,  Cor.  i).     Then  CE  is  ||  to  L. 

For  C  and  E  are  the  middle  points  of  two  sides  of  a  triangle 

of  which  L  is  the  base.  (84°,  Cor.  2) 

121°.  The  Square. — The  square  consists  of  two  rules  with 
their  edges  fixed  permanently  at 
right  angles,  or  of  a  triangular 
plate   of  wood  or  metal  having 

Dtwo  of  its  edges  at  right  angles. 
To  test  a  square. 
Draw  a  line  AB  and  place  the 
square  as  at  S,  so  that  one  edge 
coincides  with  the  line,  and  along 
the   other    edge    draw    the    line 
CD. 
Next  place  the  square  in  the  position  S'.     If  the  edges  can 


CONSTRUCTIVE   GEOMETRY. 


79 


now  be  made  to  coincide  with  the  two  lines  the  square 
is  true. 

This  test  depends  upon  the  fact  that  a  right  angle  is  one- 
half  a  straight  angle. 

The  square  is  employed  practically  for  drawing  a  line  J_  to 
another  line.  — 


Cor.  I.  The  square  is  employed  to 
draw  a  series  of  parallel  lines,  as  in 
the  figure. 

Cor.  2.  To  draw  the  bisectors  of  an  angle  by  means  of  the 
square. 

Let  AOB  be  the  given  angle.  Take  OA  =  OB,  and  at  A 
and  B  draw  perpendiculars  to  OA  and  OB. 

Since  AOB  is  not  a  straight  angle,  these  perpendiculars 
meet  at  some  point  C.  (79°,  Cor.) 

Then  OC  is  the  internal  bisector  of  ^lAOB.  For  the  tri- 
angles AOC  and  BOC  are  evidently  congruent. 

^AOC  =  ^BOC. 
The  line  drawn  through  O  _L  to  OC  is  the  external  bisector. 

122°.  Problem. — Through  a  given  point  in  a  line  to  draw  a 
line  which  shall  make  a  given  angle 
with  that  line. 

Let  P  be  the  given  point  in  the 
line  L,  and  let  X  be  the  given 
angle. 

Constr.—  Yxom  any  point    B 
the  arm  OB  draw  a  ±  to  the  arm  OA. 

Make  PA'  =  OA,  and  at  A'  draw  the  J.  A'B'  making 
A'B'  =  AB,     PB'  is  the  line  required. 

Proof. — The  triangles  OBA  and  PB'A'  are  evidently  con- 
gruent, and  .*.  z.BOA  =  X  =  ^B'PA'. 

Cor.  Since  PA'  might  have  been  taken  to  the  left  of  P,  the 
problem  admits  of  two  solutions.  When  the  angle  X  is  a 
right  angle  the  two  solutions  become  one. 


(120°) 


8o 


SYNTHETIC   GEOMETRY. 


123°.  The  Protractor. — This  instrument  has  different  forms 
depending  upon  the  accuracy  re- 
quired of  it.  It  usually  consists  of 
a  semicircle  of  metal  or  ivory  divided 
into  degrees,  etc.  (41°).  The  point  C 
is  the  centre.  By  placing  the  straight 
edge  of  the  instrument  in  coincidence  with  a  given  line  AB 
so  that  the  centre  falls  at  a  given  point  C,  we  can  set  off  any 
angle  given  in  degrees,  etc.,  along  the  arc  as  at  D.  Then 
the  line  CD  passes  through  C  and  makes  a  given  angle 
with  AB. 


124°.  Problem.- 


Given  the  sides  of  a  triangle  to  construct  it. 
Constr. — Place  the  three  sides  of 
the  triangle  in   line,  as  AB,  15 C, 
CD. 
With  centre  C  and  radius  CD 
D  describe  a  circle,  and  with  centre 
B  and  radius  BA  describe  a  circle. 
Let  E  be  one  point  of  intersection  of  these  circles. 
Then  A^EC  is  the  triangle  required. 

Proof.— V,^  =  V>K  and  CE  =  CD. 

Since  the  circles  intersect  in  another  point  E',  a  second 
triangle  is  formed.  But  the  two  triangles  being  congruent 
are  virtually  the  same  triangle. 

Cor.  I.  When  AB  =  BC=?  C A  the  triangle  is  equilateral. 

(53°,  Def.  2) 

In  this  case  the  circle  AE  passes  through  C  and  the  circle 
DE  through  B,  so  that  B  and  C  become  the  centres  and  BC 
a  common  radius. 

Cor.  2.  When  BC  is  equal  to  the  sum  or  difference  of  AB 
and  CD  the  circles  touch  (113°,  Def.)  and  the  triangle  takes 
the  limiting  form  and  becomes  a  line. 

When  BC  is  greater  than  the  sum  or  less  than  the  differ- 


CONSTRUCTIVE   GEOMETRY.  8 1 

ence  of  AB  and  CD  the  circles  do  not  meet  (113°,  Def.)  and 
no  triangle  is  possible. 

Therefore  that  three  line-segments  may  form  a  triangle, 
each  one  must  be  less  than  the  sum  and  greater  than  the 
difference  of  the  other  two. 


125°.  The  solution  of  a  problem  is  sometimes  best  effected 
by  supposing  the  construction  made,  and  then  by  reasoning 
backwards  from  the  completed  figure  to  some  relation 
amongst  the  given  parts  by  means  of  which  we  can  make 
the  construction. 

This  is  analogous  to  the  process  employed  for  the  solution 
of  equations  in  Algebra,  and  a  more  detailed  reference  will 
be  made  to  it  at  a  future  stage. 

The  next  three  problems  furnish  examples 

126°.  Problein. — To  construct  a  triangle  when  two  sides 
and  the  median  to  the  third  side  are  given. 

Let  a  and  b  be  two  sides  and  n 
the  median  to  the  third  side. 

Suppose  ACB  is  the  required 
triangle  having  CD  as  the  given 
median. 

By  completing  the  ^^ACBC 
and  joining    DC,   we   have   DC  c' 

equal  to  CD  and  in  the  same  line,  and  BC=AC  (81°) ;  and 
the  triangle  CCB  has  CC  =  2;/,  CB  =  rt:,  and  BC=AC  =  ^, 
and  is  constructed  by  124°. 

Thence  the  triangle  ACB  is  readily  constructed. 

Cor.  Since  CC  is  twice  the  given  median,  and  since  the 
possibility  of  the  triangle  ACB  depends  upon  that  of  CCB, 
therefore  a  median  of  a  triangle  is  less  than  one-half  the 
sum,  and  greater  than  one-half  the  difference  of  the  conter- 
minous sides.  (124°,  Cor.  2) 

F 


82 


SYNTHETIC   GEOMETRY. 


127°.  Problem. — To  trisect   a  given   line-segment,  i.e.^  to 
divide  it  into  three  equal  parts. 

Cojistr. — Let  AB  be  the  segment. 
Through  A  draw  any  line  CD  and  make 

Af •^^ 3^B  AC=AD.     Bisect  DB  in  E,  and  join  CE, 

E  cutting  AB  in  F, 

D  Then  AF  is  |AB. 

Proof. — CBD  is  a  A  and  CE  and  BA  are  two  medians. 

AF=|AB.  (85°,  Cor.) 

Bisecting  FB  gives  the  other  point  of  division. 


128°.  Problem. — To  construct  a  A  when  the  three  medians 


/, 


are  given. 

Let    /,    m,    n    be    the 

medians,     and     suppose 

to     be     the    required 

Then 

BE  =  ?«,     and  CF  =  ;2, 


given 
ABC 
triangle, 
c 

AD  = 

AO  =  f/,  OB  =  |;;z,  and  OY  =  \n, 
.'.  in  the  AAOB  we  have  two  sides  and  the  median  to  the 
third  side  given.     Thence  AAOB  is   constructed  by   126° 
and  127°. 

Then  producing  FO  until  0C  =  2F0,  C  is  the  third  vertex 
of  the  triangle  required. 

Ex.  To  describe  a  square  whose  sides  shall  pass  through 
four  given  points. 

Let  P,  Q,  R,  S  be  the  given  points,  and 
suppose  ABCD  to  be  the  square  required. 
Join  P  and  Q  upon  opposite  sides  of  the 
square,  and  draw  QG  ||  to  BC.  Draw  SX 
J_  to  PQ  to  meet  BC  in  E,  and  draw  EF 
II  to  CD.  Then  AQPG  =  AFSE, 
and  SE  =  PO. 

Hence  the  construction  : — 
Join  any  two  points  PQ,  and  through  a  third  point  S  draw 


CONSTRUCTIVE   GEOMETRY.  S^ 

SX  ±  to  PQ.  On  SX  take  SE  =  PQ  and  join  E  with  the 
fourth  point  R.  ER  is  a  side  of  the  square  in  position  and 
direction,  and  the  points  first  joined,  P  and  Q,  are  on  oppo- 
site sides  of  the  required  square. 

Thence  the  square  is  readily  constructed. 

Since  SE  may  be  measured  in  two  directions  along  the 
line  SX,  two  squares  can  have  their  sides  passing  through 
the  same  four  points  P,  Q,  R,  S,  and  having  P  and  Q  on 
opposite  sides. 

Also,  since  P  may  be  first  connected  with  R  or  S,  two 
squares  can  be  constructed  fulfilling  the  conditions  and 
having  P  and  Q  on  adjacent  sides. 

Therefore,  four  squares  can  be  constructed  to  have  their 
sides  passing  through  the  same  four  given  points. 


CIRCLES  FULFILLING  GIVEN  CONDITIONS. 

The  problems  occurring  here  are  necessarily  of  an  elemen- 
tary character.  The  more  complex  problems  require  relations 
not  yet  developed. 

129°.  Problem. — To  describe  a  circle  to  touch  a  given  line 
at  a  given  point.  m 

P  is  a  given  point  in  the  line  L. 

Co7istr. — Through  P  draw  M  J_  to  L. 
A  circle  having  any  point  C,  on  M,  as 
centre  and  CP  as  radius  touches  L  at  P. 

Proof. — L  is  ±  to  the  diameter  at  its 
end-point,  therefore  L  is  tangent  to  the  circle.    (110°,  Cor.  3) 

Def. — As  C  is  a7iy  point  on  M,  any  number  of  circles  may 
be  drawn  to  touch  L  at  the  point  P,  and  all  their  centres  lie 
on  M. 

Such  a  problem  is  indefinite  because  the  conditions  are 
not  sufficient  to  determine  a  particular  circle.     If  the  circle 


84  SYNTHETIC   GEOMETRY. 

varies  its  radius  while  fulfilling  the  conditions  of  the  problem, 
the  centre  moves  along  M  ;  and  M  is  called  the  centre-locus 
of  the  variable  circle. 

Hence  the  centre-locus  of  a  circle  which  touches  a  fixed 
line  at  a  fixed  point  is  the  perpendicular  to  the  line  at  that 
point. 

Cor.  If  the  circle  is  to  pass  through  a  second  given  point 
Q  the  problem  is  definite  and  the  circle  is  a  particular  one, 
since  it  then  passes  through  three  fixed  points,  viz.,  the  double 
point  P  and  the  point  Q.  (109°)  4) 

In  this  case  Z.COP=Z_CPQ. 

But  ^CPQ  is  given,  since  P,  Q,  and  the  line  L  are  given. 
.*.  z_CQP  is  given  and  C  is  a  fixed  point. 

130°.  Problem. — To  describe  a  circle  to  touch  two  given 
non-parallel  lines. 

Let  L  and  M  be  the  lines  inter- 
secting at  O. 

Draw  N,  N,  the  bisectors  of  the 
angle  between  L  and  M.(i2i°,  Cor.  2) 
From  C,  any  point  on  either  bi- 
sector, draw  CA  X  to  L. 
The  circle  with  centre  C  and  radius  CA  touches  L,  and  if 
CB  be  drawn  X  to  M,  CB  =  CA.  (68°) 

Therefore  the  circle  also  touches  M. 

As  C  is  any  point  on  the  bisectors  the  problem  is  indefinite, 
and  the  centre-locus  of  a  circle  which  touches  two  intersecting 
lines  is  the  two  bisectors  of  the  angle  between  the  lines . 

131°.  Prohle77i. — To  describe  a  circle  to  touch  three  given 
lines  which  form  a  triangle. 

L,  M,  N  are  the  lines  forming  the  triangle . 

Constr. — Draw  Ij,  E^,  the  internal  and  external  bisectors 
of  the  angle  A ;  and  I2,  Eg,  those  of  the  angle  B. 

^A  +  /.Bis<J_,   .-.  ^BAO-fiLABO  is<~l- 


CONSTRUCTIVE  GEOMETRY. 


8S 


.*.  Ii  and  I2  meet  at  some  point  O  (79°)  and  are  not  X  to  one 
another  and  therefore  Ej  and  Eg  meet  at  some  point  O3. 

m\        /m  (79°,  Cor.) 


Also  Ii  and  Eg  meet  at  some  point  Oj,  and  similarly  Ig  and 
El  meet  at  O2. 

The  four  points  O,  O^,  Og,  O3  are  the  centres  of  four  circles 
each  of  which  touches  the  three  lines  L,  M,  and  N. 

Proof. — Circles  which  touch  M  and  N  have  \  and  E^  as 
their  centre-locus  (130°),  and  circles  which  touch  N  and  L 
have  Ig  and  Eg  as  their  centre-locus. 

.'.  Circles  which  touch  L,  M,  and  N  must  have  their 
centres  at  the  intersections  of  these  loci. 

But  these  intersections  are  O,  O^,  Og,  and  O3, 
.*.  O,  Oi,  Og,  and  O3  are  the  centres  of  the  circles  required. 

The  radii  are  the  perpendiculars  from  the  centres  upon  any 
one  of  the  lines  L,  M,  or  N. 

Cor.  I.  Let  I3  and  E3  be  the  bisectors  of  the  Z.C.  Then, 
since  O  is  equidistant  from  L  and  M,  I3  passes  through  O.  (68°) 

.*.  the  three  internal  bisectors  of  the  angles  of  a  triangle 
are  concurrent. 

Cor.  2.  Since  O3  is  equidistant  from  L  and  M,  I3  passes 
through  O3.  (68°) 


86  SYNTHETIC   GEOMETRY. 

.*.  the  external  bisectors  of  two  angles  of  a  triangle  and  the 
internal  bisector  of  the  third  angle  are  concurrent. 

Def.  I. — When  three  or  more  points  are  in  line  they  are 
said  to  be  collinear. 

Cor.  3.  The  line  through  any  two  centres  passes  through  a 

vertex  of  the  AABC. 

.*.  any  two  centres  are  collinear  with  a  vertex  of  the  A- 
The  lines  of  collinearity  are  the  six  bisectors  of  the  three 

angles  A,  B,  and  C. 

Def.  2  — With  respect  to  the  AABC,  the  circle  touching 
the  sides  and  having  its  centre  at  O  is  called  the  i7iscribed 
circle  or  simply  the  in-circle  of  the  triangle. 

The  circles  touching  the  lines  and  having  centres  at  Oj, 
O2,  and  O3  are  the  escribed  or  ex-circles  of  the  triangle. 


REGULAR  POLYGONS. 

132°.  Def.  I. — A  closed  rectilinear  figure  without  re-entrant 
angles  (89°,  2)  is  in  general  called  z. polygon. 

They  are  named  according  to  the  number  of  their  sides  as 
follows  : — 

3,  triangle  or  trigon  ; 

4,  quadrangle,  or  tetragon,  or  quadrilateral ; 

5,  pentagon  ;  6,  hexagon  ;     7,  heptagon  ; 

8,  octagon  ;    10,  decagon  ;  12,  dodecagon  ;  etc. 

The  most  important  polygons  higher  than  the  quadrangle 
are  regular  polygons. 

Def  2. — A  regular  polygon  has  its  vertices  concyclic,  and 
all  its  sides  equal  to  one  another. 

The  centre  of  the  circumcircle  is  the  centre  of  the  polygon. 

133".  Theorem. — If  ;/  denotes  the  number  of  sides  of  a 


CONSTRUCTIVE   GEOMETRY.  8/ 

regular  polygon,  the  magnitude  of   an  internal    angle    is 

(  2  -  4  J  right  angles. 
\       nf 

Proof. — Let  AB,  BC  be  two  consecu- 
tive sides  of  the  polygon  and  O  its  centre. 
Then   the  triangles  AOB,  BOG   are 
isosceles  and  congruent. 

^OAB=^OBA-/.OBC  =  etc., 
^OAB  +  ^OBA  =  /.ABC. 
But  ^OAB  +  ^OBA-J_-zAOB, 

and  (132°,  Def.  2)  ^ob  =  4  right  angles^ 


z.ABC  =  (2-4)  right  angles, 


i^-)} 


90" 


q.e.d. 


Cor.  The  internal  angles  of  the  regular  polygons  expressed 
in  right  angles  and  in  degrees  are  found,  by  putting  proper 
values  for  «,  to  be  as  follows  : — 

Equilateral  triangle,  f      60°         Octagon,      .  f     135° 

Square, i       90°         Decagon,     .  f     144° 

Pentagon,  .     .     .     .  f     108°         Dodecagon,    f     150° 
Hexagon,    .     .     .     .  ^     120° 

134°.  Problem. — On  a  given  line-segment  as  side  to  con- 
struct a  regular  hexagon.  A^ 
Let  AB  be  the  given  segment. 

Constr. — On   AB    construct   the  equi-  pj 
lateral  triangle  AOB  (124°,  Cor.  i),  and 
with  O  as  centre  describe  a  circle  through 
A,  cutting  AO  and  BO  produced  in  D  e." 

and  E.     Draw  FC,  the  internal  bisector  of  Z.AOE. 
ABCDEF  is  the  hexagon. 

Proof.—  ^AOB  =  ^EOD  =  O 

Z.A0E=O  and  A0F  =  O 
^A0B  =  /.B0C  =  ^C0D  =  etc.  =  |n. 


88 


SYNTHETIC   GEOMETRY. 


And  the  chords  AB,  BC,  CD,  etc.,  being  sides  of  congruent 
equilateral  triangles  are  all  equal. 

Therefore  ABCDEF  is  a  regular  hexagon. 

Cor.  Since  AOB  is  an  equilateral  triangle,  AB=AO  ; 
.*.  the  side  of  a  regular  hexagon  is  equal  to  the  radius  of 
its  circumcircle. 


135°.  Pi'oblem. — To  determine  which  species  of  regular 
polygons,  each  taken  alone,  can  fill  the  plane. 

That  a  regular  polygon  of  any  species  may  be  capable  of 
filling  the  plane,  the  number  of  right  angles  in  its  internal 
angle  must  be  a  divisor  of  4.  But  as  no  internal  angle  can 
be  so  great  as  two  right  angles,  the  only  divisors,  in  133°, 
Cor.,  are  f,  i,  and  |,  which  give  the  quotients  6,  4,  and  3. 

Therefore  the  plane  can  be  filled  by  6  equilateral  triangles, 
or  4  squares,  or  3  hexagons. 

It  is  worthy  of  note  that,  of  the  three  regular  polygons 
which  can  fill  the  plane,  the  hexagon  includes  the  greatest 
area  for  a  given  perimeter.     As  a  consequence,  the  hexagon 

is  frequently  found  in 
Nature,  as  in  the  cells  of 
bees,  in  certain  tissues  of 
plants,  etc. 

Ex.  I.  Let  D,  E,  F  be 
points  of  contact  of  the  in- 
circle,  and  P,  P',  P",  R,  R', 
R",  etc.,  of  the  ex- circles. 

(131") 
ThenAP  =  AP',CP'  =  CP", 
andBP  =  BP",(ii4°,Cor.i) 
.-.  AP'  +  AP 

=AB+BC+AC 

and,  denoting  the  perimeter  of  the  triangle  by  is^  we  have 
AP  =  AF=J, 


CONSTRUCTIVE   GEOMETRY.  89 

CP'  =  J'-^  =  CP",      BP=j-r=BP". 
Similarly,       AR  =  ^-^  =  AR",    BR'  =  s-a  =  BR",  etc. 
Again,  CD  =  CE  =  ^-AE  =  ^-AF  =  ^-(^-BF) 

CE  =  CD=j-^=BP". 
Similarly,      AE  =  AF  =  j  -  ^  =  BR",  etc. 
These  relations  are  frequently  useful. 

If  we  put  Az  to  denote  the  distance  of  the  vertex  A  from 
the  adjacent  points  of  contact  of  the  in-circle,  and  A^,  Ac  to 
denote  its  distances  from  the  points  of  contact  of  the  ex-circles 
upon  the  sides  d  and  c  respectively,  we  have 
A/=  B^  =  C  <^ = i- -  ^'-r, 
Bi  =  Ca  =  Ac=s-d, 
Cz=Ad=Ba  =  s  —  c. 


Exercises. 

1.  In  testing  the  straightness  of  a  "  rule "  three  rules  are 

virtually  tested.     How  ? 

2.  To  construct  a  rectangle,  and  also  a  square. 

3.  To  place  a  given  line-segment  between  two  given  lines 

so  as  to  be  parallel  to  a  given  line. 

4.  On  a  given  line  to  find  a  point  such  that  the  lines  joining 

it  to  two  given  points  may  make  equal  angles  with  the 
given  line. 

5.  To  find  a  point  equidistant  from  three  given  points. 

6.  To  find  a  line  equidistant  from  three  given  points.     How 

many  lines  ? 

7.  A  is  a  point  on  line  L  and  B  is  not  on  L.    To  find  a  point 

P  such  that  PA±PB  may  be  equal  to  a  given  segment. 

8.  On  a  given  line  to  find  a  point  equidistant  from  two 

given  points. 

9.  Through  a  given  point  to  draw  a  line  which  shall  form  an 

isosceles  triangle  with  two  given  lines.     How  many 
solutions  ? 


90  SYNTHETIC   GEOMETRY. 

10.  Through  two  given  points  on  two  parallel  lines  to  draw 

two  lines  so  as  to  form  a  rhombus. 

11.  To  construct  a  square  having  one  of  its  vertices  at  a 

given  point,  and  two  other  vertices  lying  on  two  given 
parallel  lines. 

12.  Through  a  given  point  to  draw  a  line  so  that  the  intercept 

between  two  given  parallels  may  be  of  a  given  length. 

13.  To  construct  a  triangle  when  the  basal  angles  and  the 

altitude  are  given. 

14.  To  construct  a  right-angled  triangle  when  the  hypothen- 

use  and  the  sum  of  the  sides  are  given. 

1 5.  To  divide  a  line-segment  into  any  number  of  equal  parts. 

16.  To  construct  a  triangle  when  the  middle  points  of  its 

sides  are  given. 

17.  To  construct  a  parallelogram  when  the  diagonals  and 

one  side  are  given. 

18.  Through  a  given  point  to  draw  a  secant  so  that  the  chord 

intercepted  by  a  given  circle  may  have  a  given  length. 

19.  Draw  a  line  to  touch  a  given  circle  and  be  parallel  to  a 

given  line.     To  be  perpendicular  to  a  given  line. 

20.  Describe  a  circle  of  given  radius  to  touch  two  given  lines. 

21.  Describe  a  circle  of  given  radius  to  touch  a  given  circle 

and  a  given  line. 

22.  Describe  a  circle  of  given  radius  to  pass  through  a  given 

point  and  touch  a  given  circle. 

23.  Describe  a  circle  of  given  radius  to  touch  two  given  circles. 

24.  To  inscribe  a  regular  octagon  in  a  circle. 

25.  To  inscribe  a  regular  dodecagon  in  a  circle. 

26.  A,  B,  C,  D,  ...,  are  consecutive  vertices   of  a  regular 

octagon,  and  A,  B',  C,  D',  ...,  of  a  regular  dodecagon 
in  the  same  circle.  Find  the  angles  between  AC  and 
B'C;  between  BE'  and  B'E.     (Use  io8°.) 

27.  Show  that  the  plane  can  be  filled  by 

{a)  Equilateral  triangles  and  regular  dodecagons. 
{b)  Equilateral  triangles  and  squares. 
{c)  Squares  and  regular  octagons. 


PART    II 


PRELIMINARY. 

136°.  Def.  I. — The  area  of  a  plane  closed  figure  is  the  por- 
tion of  the  plane  contained  within  the  figure,  this  portion 
being  considered  with  respect  to  its  extent  only,  and  without 
respect  to  form. 

A  closed  figure  of  any  form  may  contain  an  area  of  any 
given  extent,  and  closed  figures  of  different  forms  may  con- 
tain areas  of  the  same  extent,  or  equal  areas. 

Def.  2. — Closed  figures  are  equal  to  one  another  when  they 
include  equal  areas.  This  is  the  definition  of  the  term 
"  equal "  when  comparing  closed  figures. 

Congruent  figures  are  necessarily  equal,  but  equal  figures 
are  not  necessarily  congruent.  Thus,  a  A  and  a  [=3  may  have 
equal  areas  and  therefore  be  equals  although  necessarily 
having  different  forms. 

137°.  Areas  are  compared  by  superposition.  If  one  area 
can  be  superimposed  upon  another  so  as  exactly  to  cover  it, 
the  areas  are  equal  and  the  figures  containing  the  areas  are 
equal.  If  such  superposition  can  be  shown  to  be  impossible 
the  figures  are  not  equal. 

In  comparing  areas  we  may  suppose  one  of  them  to 
be  divided  into  any  requisite  number  of  parts,  and  these 
parts  to  be  afterwards  disposed  in  any  convenient  order, 
since  the  whole  area  is  equal  to  the  sum  of  all  its  parts. 

91 


92  SYNTHETIC   GEOMETRY. 

Illustration. — ABCD  is  a  square. 

Then  the  AABC^AADC,  and  they  are  therefore  equal. 

6 ^  Now,  if  AD  and  DE  be  equal  and  in 

line,  the  As  ADC  and  EDC  are  con- 
gruent and  equal. 

Therefore  the  AABC  may  be  taken 
"e  from  its  present  position  and  be  put  into 
the  position  of  CDE.     And  the  square  ABCD  is  thus  trans- 
formed into  the  A  ACE  without  any  change  of  area  ; 
QABCD^AACE. 
It  is  evident  that  a  plane  closed  figure  may  be  considered 
from  two  points  of  view. 

1.  With  respect  to  the  character  and  disposition  of  the 
lines  which  form  it.  When  thus  considered,  figures  group 
themselves  into  triangles,  squares,  circles,  etc.,  where  the 
members  of  each  group,  if  not  of  the  same  form,  have  at 
least  some  community  of  form  and  character. 

2.  With  respect  to  the  areas  enclosed. 

When  compared  from  the  first  point  of  view,  the  capability 
of  superposition  is  expressed  by  saying  that  the  figures  are 
congruent.  When  compared  from  the  second  point  of  view, 
it  is  expressed  by  saying  that  the  figures  are  equal. 

Therefore  congruence  is  a  kind  of  higher  or  double 
equality,  that  is,  an  equality  in  both  form  and  extent  of  area. 
This  is  properly  indicated  by  the  triple  lines  (  =  )  for  con- 
gruence, and  the  double  lines  (  =  )  for  equality. 

138°.  Def. — The  altitude  of  a  figure  is  the  line-segment 
which  measures  the  distance  of  the  farthest  point  of  a  figure 
from  a  side  taken  as  base. 

The  terms  base  and  altitude  are  thus  correlative.  A  tri- 
angle may  have  three  different  bases  and  as  many  corre- 
sponding altitudes.  ^1°) 

In  the  rectangle  (82°,  Def.  2)  two  adjacent  sides  being 
perpendicular  to  one  another,  either  one  may  be  taken  as  the 


COMPARISON    OF   AREAS.  93 

base  and  the  adjacent  one  as  the  altitude.  The  rectangle 
having  two  given  segments  as  its  base  and  altitude  is  called 
the  rectangle  on  these  segments. 

Notation. — The  symbol  im  stands  for  the  word  rectangle 
and  I      7  iox  parallelogram. 

Rectangles  and  parallelograms  are  commonly  indicated  by 
naming  a  pair  of  their  opposite  vertices. 


SECTION  I. 

COMPARISON    OF   AREAS—RECTANGLES, 
PARALLELOGRAMS,   TRIANGLES. 

139°.  Theorem. — i.  Rectangles  with  equal  bases  and  equal 
altitudes  are  equal. 

2.  Equal  rectangles  with  equal  bases  have  equal  altitudes. 

3.  Equal  rectangles  with  equal  altitudes  have  equal  bases.^ 

c 
Q 

D 
G 


I.  In  the  OS  BD  and  FH,  if  \ 
AD  =  EH, 

and  AB  =  EF,  ^ 

then  [=]BD=c=iFH.  ^ 


Proof. — Place   E   at  A  and   EH    along 
AD.     Then,  as  ^FEH  =z.BAD  ="],  EF  will  lie  along  AB. 

And  because  EH  =  AD  and  EF  =  AB,  therefore  H  falls  at 
D  and  F  at  B,  and  the  two  cds  are  congruent  and  therefore 
equal.  q.e.d. 

2.  If[=iBD  =  i=iFH  andAD  =  EH,  then  AB  =  EF. 

Proof.— \{  EF  is  not  equal  to  AB,  let  AB  be  >  EF. 
Make  AP  =  EF  and  complete  the  cuPD. 
Then  oP  D  =  aF  H ,  by  the  first  part, 

but  i=iBD=aFH,  (hyp.) 


94  SYNTHETIC   GEOMETRY. 

i=iPD  =  [=iBD,  which  is  not  true, 
.*.  AB  and  EF  cannot  be  unequal,  or 

AB  =  EF.  q,e.d. 

3.  lf[=iBD=[=iFH  and  AB  =  EF,  then  AD  =  EH. 

Proof.— l^et  AB  and  EF  be  taken  as  bases  and  AD  and 

EH  as  altitudes   (138°),  and  the  theorem  follows  from  the 

second  part.  q.e.d. 

Cor.  In  any  rectangle  we  have  the  three  parts,  base,  alti- 
tude, and  area.  If  any  two  of  these  are  ^iven  the  third  is 
given  also. 

140°.   Theorem. — A  parallelogram  is  equal  to  the  rectangle 

E .B F c  on  its  base  and  altitude. 

/        AC  is  a  I      7  whereof  AD  is  the  base 
and  DF  is  the  altitude. 

Then  nZ7AC=i=i  on  AD  and  DF. 

/•r^^/— Complete  the  cuADFE  by  drawing  AE  ±  to  CB 
produced. 

Then  AAEB  =  ADFC,    '.•  AE  =  DF,  AB  =  DC, 

and  ^EAB  =  ^FDC; 

.*.  ADFC    may  be  transferred  to  the  position  AEB,  and 
^ZIjABCD  becomes  the  cuAEFD, 

ZZZ7AC  =  [=i  on  AD  and  DF.  q.e.d. 

Cor.  I .  Parallelograms  with  equal  bases  and  equal  altitudes 
are  equal.     For  they  are  equal  to  the  same  rectangle. 

Cor.  2.  Equal  parallelograms  with  equal  bases  have  equal 
altitudes,  and  equal  parallelograms  with  equal  altitudes  have 
equal  bases. 

Cor.  3.  If  equal  parallelograms  be  upon  the  same  side  of 
the  same  base,  their  sides  opposite  the  common  base  are 
in  line. 

141°.  Theorem. — A  triangle  is  equal  to  one-half  the  rect-. 
angle  on  its  base  and  altitude. 


COMPARISON   OF  AREAS. 


95 


ABC  is  a  triangle  of  which  AC  is  the  base  and  BE  the 

altitude.  b  P 

Then  AABC  =  |(=i  on  AC  and  BE. 

Pr^^/— Complete  the  £I^ABDC,  of 
which  AB  and  AC  are  adjacent  sides,    a 
Then  AABC  =  ADCB, 

AABC=|£ZI7AD=|i=]  on  AC  and  BE.    (140°)  q.e.d. 

Cor.  I.  A  triangle  is  equal  to  one-half  the  parallelogram 
having  the  same  base  and  altitude. 

Cor.  2.  Triangles  with  equal  bases  and  equal  altitudes  are 
equal.     For  they  are  equal  to  one-half  of  the  same  rectangle. 

Cor.  3.  A  median  of  a  triangle  bisects  the  area.  For  the 
median  bisects  the  base. 

Cor.  4.  Equal  triangles  with  equal  bases  have  equal  alti- 
tudes, and  equal  triangles  with  equal  altitudes  have  equal  bases. 

Cor.  5.  If  equal  triangles  be  upon  the  same  side  of  the 
same  base,  the  line  through  their  vertices  is  parallel  to  their 
common  base. 


142°.   Theorem. — If  two  triangles  are  upon  opposite  sides 
of  the  same  base — 

1.  When  the  triangles  are  equal,  the  base  bisects  the  seg- 
ment joining  their  vertices  ; 

2.  When  the  base  bisects  the  segment  joining  their  vertices, 
the  triangles  are  equal.    (Converse  of  i.) 

ABC  and  ADC  are  two  triangles  upon 
opposite  sides  of  the  common  base  AC. 

A' 


I.  If 

then 


AABC  =  AADC, 
BH  =  HD. 


Proof.— Lq\.  be  and  DF  be  altitudes, 

Then   •.*  AABC  =  AADC,      .'.  BE  =  DF, 

AEBH=AFDH,  andBH  =  HD. 

2.  IfBH  =  HD,  then  AABC=AADC. 


q.e.d. 


X  Y 


96  SYNTHETIC   GEOMETRY. 

Proof.— Since  BH  =  HD,      .'.  AABH=AADH, 
and  ACBH=ACDH.  (141°,  Cor.  3) 

/.adding,       AABC=AADC.  q.e.d. 

143.  Def. — By  the  stem  or  difference  of  two  closed  figures  is 
meant  the  sum  or  difference  of  the  areas  of  the  figures. 

If  a  rectangle  be  equal  to  the  sum  of  two  other  rectangles 
its  area  may  be  so  superimposed  upon  the  others  as  to  cover 
both. 

144°.  Theorem. — If  two  rectangles  have  equal  altitudes, 
their  sum  is  equal  to  the  rectangle  on  their  common  altitude 
and  the  sum  of  their  bases. 

'^ D F       Proof.— Lei  the  as  X  and  Y,  having 

equal  altitudes,  be  so  placed  as  to  have 
their  altitudes  in  common  at  CD,  and  so 
that  one  en  may  not  overlap  the  other. 
Then  ^BDC=z.CDF="]» 

BDF  is  a  Hne.  (38°,  Cor.  2) 

Similarly  ACE  is  a  line. 

But  BD  is  II  to  AC,  and  BA  is  ||  to  DC  ||  to  FE  ;  therefore 
AF  is  the  m  on  the  altitude  AB  and  the  sum  of  the  bases 
AC  and  CE  ;  and  the  [=]AF=cdAD  +0CF.  q.e.d. 

Cor.  I.  If  two  triangles  have  equal  altitudes,  their  sum  is 
equal  to  the  triangle  having  the  same  altitude  and  having  a 
base  equal  to  the  sum  of  the  bases  of  the  two  triangles. 

Cor.  2.  If  two  triangles  have  equal  altitudes,  their  sum  is 
equal  to  one-half  the  rectangle  on  their  common  altitude 
and  the  sum  of  their  bases. 

Cor.  3.  If  any  number  of  triangles  have  equal  altitudes, 
their  sum  is  equal  to  one-half  the  rectangle  on  their  common 
altitude  and  the  sum  of  their  bases. 

In  any  of  the  above,  "base"  and  "altitude"  are  inter- 
changeable. 


COMPARISON    OF   AREAS.  .  97 

145°.  Theorem. — Two  lines  parallel  to  the  sides  of  a 
parallelogram  and  intersecting  upon  a  diagonal  divide  the 
parallelogram  into  four  parallelograms  such  that  the  two 
through  which  the  diagonal  does  not  pass  are  equal  to  one 
another. 

In  the£IZ7ABCD,EFis  ||  to  AD  and 
GH  is  II  to  BA,  and  these  intersect  at 
O  on  the  diagonal  AC. 

Then        /=Z7BO  =  Z=Z70D. 

/'r^^/— AABC  =  AADC,  and  AAEO=AAHO, 
and  AOGC  =  AOFC  ;  (141°,  Cor.  i) 

but  £IZ7BO=AABC-AAEO-AOGC, 

and  C=70D=AADC-AAH0-A0FC. 

'B0  =  /=Z70D.  q.e.d. 


Cor.  I.  £Z7BF  =  ^=7GD. 

Cor.  2.  If  £Z:7B0  =  £Z70D,  O  is  on  the   diagonal  AC. 
(Converse  of  the  theorem.) 

For  if  O  is  not  on  the  diagonal,  let  the  diagonal  cut  EF  in 
O'.     Then  Z=II7BO'  =  iCZ70'D.  (145°) 

But  £ZI7B0'  is  <  ^IZ7BO,  and  .CZJO'D  is  >  ZZZ70D  ; 
.'.  I      7BO  is  >  I      7OD,  which  is  contrary  to  the  hypothesis; 
.'.  the  diagonal  cuts  EF  in  O. 

Ex.  Let  ABCD  be  a  trapezoid. 
(84°,  Def )  In  line  with  AD  make 
DE  =  BC,  and  in  line  with  BC  make 
CF-AD. 

Then  BF  =  AE  and  BFEA  is  a 


EX 


But  the  trapezoid  CE  can  be  superimposed  on  the  trape- 
zoid DB,  since  the  sides  are  respectively  equal,  and 
^F=A,  and  :lE  =  B,  etc. 
trapezoid  BD=|/^I7BE, 
or,  a  trapezoid  is  equal  to  one-half  the  rectangle  on  its  alti- 
tude and  the  sum  of  its  bases. 

G 


98  SYNTHETIC   GEOMETRY. 

Exercises. 

1.  To  construct  a  triangle  equal  to  a  given  quadrangle. 

2.  To  construct  a  triangle  equal  to  a  given  polygon. 

3.  To  bisect  a  triangle  by  a  line  drawn  through  a  given 

point  in  one  of  the  sides. 

4.  To  construct  a  rhombus  equal  to  a  given  parallelogram, 

and  with  one   of  the  sides  of  the  parallelogram  as 
its  side. 

5.  The  three  connectors  of  the  middle  points  of  the  sides 

of  a    triangle    divide   the  triangle   into    four    equal 
triangles. 
6    Any  line  concurrent  with  the  diagonals  of  a  parallelogram 
bisects  the  parallelogram. 

7.  The  triangle  having  one  of  the  non-parallel  sides  of  a 

trapezoid  as  base  and  the  middle  point  of  the  opposite 
side  as  vertex  is  one-half  the  trapezoid. 

8.  The  connector  of  the  middle  points  of  the  diagonals  of  a 

quadrangle  is  concurrent  with  the  connectors  of  the 
middle  points  of  opposite  sides. 

9.  ABCD  is  a  parallelogram  and  O  is  a  point  within.    Then 

AA0B+AC0D=^ZIZ7. 
What  does  this  become  when  O  is  without  ? 
TO.  ABCD  is  a  parallelogram  and  O  is  a  point  within.    Then 
AAOC  =  AAOD-AAOB. 
What  does  this  become  when  O  is  without?    (This 
theorem  is  important  in  the  theory  of  Statics.) 

11.  Bisect  a  trapezoid  by  a  line  through  the  middle  point  of 

one   of  the   parallel  sides.      By  a   line   through   the 
middle  point  of  one  of  the  non-parallel  sides. 

12.  The  triangle  having  the  three  medians  of  another  tri- 

angle  as   its    sides    has    three-fourths    the    area    of 
the  other. 


COMPARISON    OF   AREAS. 


99 


POLYGON  AND   CIRCLE. 


146°.  Def. — The  sum  of  all  the  sides  of  a  polygon  is  called 
its  perimeter^  and  when  the  polygon  is  regular  every  side  is 
at  the  same  distance  from  the  centre.  This  distance  is  the 
apothem  of  the  polygon. 

Thus  if  ABC D... LA  be  a  regu- 
lar polygon  and  O  the  centre 
(132°,  Def.  2),  the  triangles  OAB, 
OBC,  ...   are  all  congruent,  and  Lj 

OP  =  OQ  =  etc.  o 

AB  +  BC  +  CD  +  ...  +  LA  is  the  perimeter  and  OP,  per- 
pendicular upon  AB,  is  the  apothem. 


147°.  Theorem. — A  regular  polygon  is  equal  to  one-half  the 
rectangle  on  its  apothem  and  perimeter. 

Proof.— T\i^  triangles  AOB,  BOC,  ...  LOA  have  equal 
altitudes,  the  apothem  OP,  .'.  their  sum  is  one-half  the  cu  on 
OP  and  the  sum  of  their  bases  AB  + BC-f... LA.  (144°,  Cor.  3) 

But  the  sum  of  the  triangles  is  the  polygon,  and  the  sum  of 
their  bases  is  the  perimeter. 

.*.  a  regular  polygon  =  |[Z]  on  its  apothem  and  perimeter. 


148°.  Of  a  limit. — A  limit  or  limiting  value  of  a  variable 
is  the  value  to  which  the  variable  by  its  variation  can  be 
made  to  approach  indefinitely  near, 
but  which  it  can  never  be  made  to 
pass. 

Let  ABCD  be  a  square  in  its  cir- 
cumcircle.  If  we  bisect  the  arcs  AB, 
BC,  CD,  and  DA  in  E,  F,  G,  and  H, 
we  have  the  vertices  of  a  regular 
octagon  AEBFCGDHA.  Now,  the 
area  of  the  octagon  approaches  nearer 
to  that  of  the  circle  than  the  area  of  the  square  does ;  and  the 


lOO  SYNTHETIC   GEOxMKTRY. 

perimeter  of  the  octagon  approaches  nearer  to  the  length  of 
the  circle  than  the  perimeter  of  the  square  does  ;  and  the 
apothem  of  the  octagon  approaches  nearer  to  the  radius  of 
the  circle  than  the  apothem  of  the  square  does. 

Again,  bisecting  the  arcs  AE,  EB,  BF,  etc.,  in  I,  J,  K,  etc., 
we  obtain  the  regular  polygon  of  i6  sides.  And  all  the  fore- 
going parts  of  the  polygon  of  i6  sides  approach  nearer  to 
the  corresponding  parts  of  the  circle  than  those  of  the 
octagon  do. 

It  is  evident  that  by  continually  bisecting  the  arcs,  we  may 
obtain  a  series  of  regular  polygons,  of  which  the  last  one  may 
be  made  to  approach  the  circle  as  near  as  we  please,  but  that 
however  far  this  process  is  carried  the  final  polygon  can  never 
become  greater  than  the  circle,  nor  can  the  final  apothem 
become  greater  than  the  radius. 

Hence  the  circle  is  the  limit  of  the  perimeter  of  the  regular 
polygon  when  the  number  of  its  sides  is  endlessly  increased, 
and  the  area  of  the  circle  is  the  limit  of  the  area  of  the  poly- 
gon, and  the  radius  of  the  circle  is  the  limit  of  the  apothem 
of  the  polygon  under  the  same  circumstances. 

149°.  Theorem. — A  circle  is  equal  to  one-half  the  rectangle 
on  its  radius  and  a  line-segment  equal  in  length  to  the  circle. 

Proof. — The  0  is  the  limit  of  a  regular  polygon  when  the 
number  of  its  sides  is  endlessly  increased,  and  the  radius  of 
the  0  is  the  limit  of  the  apothem  of  the  polygon. 

But,  whatever  be  the  number  of  its  sides,  a  regular  polygon 
is  equal  to  one-half  the  CD  on  its  apothem  and  perimeter.  (147°) 

.'.  a  0  is  equal  to  one-half  the  [=]  on  its  radius  and  a  line- 
segment  equal  to  its  circumference. 

Exercises. 

I.  Show  that  a  regular  polygon  may  be  described  about  a 
circle,  and  that  the  limit  of  its  perimeter  when  the 
number  of  its  sides  is  increased  indefinitely  is  the 
circumference  of  the  circle. 


MEASUREMENT  OF  LENGTHS  AN'B  AREA^:    lOI 

2.  The  difference  between  the  areas  of  tw.a  re^u-ai:  polygo-ns^ 

one  inscribed  in  a  circle  and  the  other  circumscribed 
about  it,  vanishes  at  the  limit  when  the  number  of 
sides  of  the  polygons  increases  indefinitely. 

3.  What  is  the  limit  of  the  internal  angle  of  a  regular  polygon 

as  the  number  of  its  sides  is  endlessly  increased  ? 


SECTION   11. 

MEASUREMENT  OF  LENGTHS  AND  AREAS. 

150°.  Def. — I.  That  part  of  Geometry  which  deals  with 
the  measures  and  measuring  of  magnitudes  is  Metrical 
Geometry. 

2.  To  measure  a  magnitude  is  to  determine  how  many  unit 
magnitudes  of  the  same  kind  must  be  taken  together  to  form 
the  given  magnitude.  And  the  number  thus  determined  is 
called  the  measure  of  the  given  magnitude  with  reference  to 
the  unit  employed.  This  number  may  be  a  whole  or  a  frac- 
tional number,  or  a  numerical  quantity  which  is  not  arith- 
metically expressible.  The  word  "number"  will  mean  any 
of  these. 

3.  In  measuring  length,  such  as  that  of  a  line-segment,  the 
unit  is  a  segment  of  arbitrary  length  called  the  unit-length. 
In  practical  work  we  have  several  such  units  as  an  inch,  a 
foot,  a  mile,  a  metre,  etc.,  but  in  the  Science  of  Geometry  the 
unit-length  is  quite  arbitrary,  and  results  obtained  through  it 
are  so  expressed  as  to  be  independent  of  the  length  of  the 
particular  unit  employed. 

4.  In  measuring  areas  the  unit  magnitude  is  the  area  of  the 
square  having  the  unit-length  as  its  side.  This  area  is  the 
unit-area.     Hence  the  unit-length  and  unit-area  are  not  both 


102'  ■    '       SYNTHETIC  GEOMETRY. 

kWLt^ry,,  cforl  if'  tiiher  is  fixed  the  other  is  fixed  also,  and 
determinable. 

This  relation  between  the  unit-length  and  the  unit-area  is 
conventional,  for  we  might  assume  the  unit-area  to  be  the 
area  of  any  figure  which  is  wholly  determined  by  a  single 
segment  taken  as  the  unit-length :  as,  for  example,  an  equi- 
lateral triangle  with  the  unit-length  as  side,  a  circle  with  the 
unit-length  as  diameter,  etc.  The  square  is  chosen  because 
it  offers  decided  advantages  over  every  other  figure. 

For  the  sake  of  conciseness  we  shall  symbolize  the  term 
unit-length  by  ?/./.  and  unit-area  by  z^.a. 

5.  When  two  magnitudes  are  such  that  they  are  both 
capable  of  being  expressed  arithmetically  in  terms  of  some 
common  unit  they  are  coinviensurable^  and  when  this  is  not 
the  case  they  are  mcomme7tsiirable. 

Illus. — Let  ABCD  be  a  square,  and  let 
EF  and  HG  be  drawn  _L  to  BD,  and  EH 
and  FG  _L  to  AC.  Then  EFGH  is  a 
square  (82°,  Cor.  5),  and  the  triangles 
AEB,  APB,  BFC,  BPC,  etc.,  are  all  equal 
H         D        G      to  one  another. 

If  AB  be  taken  as  u.l.^  the  area  of  the  square  AC  is  the 
u.a.\  and  if  EF  be  taken  as  u.l.,  the  area  of  the  square 
EG  is  the  21. a. 

In  the  first  case  the  measure  of  the  square  AC  is  i,  and 
that  of  EG  is  2  ;  and  in  the  latter  case  the  measure  of  the 
square  EG  is  i,  and  that  of  AC  is  \.  So  that  in  both  cases 
the  measure  of  the  square  EG  is  double  that  of  the  square  AC. 
.'.  the  squares  EG  and  AC  are  commensurable. 
Now,  if  AB  be  taken  as  ?/./.,  EF  is  not  expressible  arith- 
metically, as  will  be  shown  hereafter. 

.".  AB  and  EF  are  incommensurable. 

151°.  Let  AB  be  a  segment  trisected  at  E  and  F  (127°), 
and  let  AC  be  the  square  on  AB.     ThenAD  =  AB.     And 


p 


1 

2 

3 

4 

'^S 

6 

7 

8 

9 

G      H 


MEASUREMENT  OF  LENGTHS  AND  AREAS.    IO3 

if  AD   be  trisected  in   the  points  K  and  M,  and  through 
E  and  F  ||s  be  drawn  to  AD,  and  through     ^     e     F 
K  and  M  ||s   be  drawn   to  AB,  the  figures 
I,  2,  3,  4,  5,  6,  7,  8,  9  are  all  squares  equal  to 
one  another. 

Now,  if  AB  be  taken  as  u.l.^  AC  is  the  u.a. ; 
and  if  AE  be  taken  as  w./.,  any  one  of  the 
small  squares,  as  AP,  is  the  u.a.  And  the  segment  AB  con- 
tains AE  3  times,  while  the  square  AC  contains  the  square 
AP  in  three  rows  with  three  in  each  row,  or  3'-^  times. 

.*.  if  any  assumed  u.l.  be  divided  into  3  equal  parts  for  a 
new  ?/./.,  the  corresponding  u.a.  is  divided  into  3^  equal  parts 
for  a  new  u.a.  And  the  least  consideration  will  show  that  this 
is  true  for  any  whole  number  as  well  as  3. 

.'.  I.  If  an  assumed  u.l.  be  divided  into  n  equal  parts  for  a 
new  u.l..,  the  corresponding  u.a.  is  divided  into  11^  equal  parts 
for  a  new  u.a.\  n  denoting  any  whole  number. 

Again,  if  any  segment  be  measured  by  the  u.l.  AB,  and 
also  by  the  21.I.  AE,  the  measure  of  the  segment  in  the  latter 
case  is  three  times  that  in  the  former  case.  And  if  any  area 
be  measured  by  the  ti.a.  AC,  and  also  by  the  ti.a.  AP,  the 
measure  of  the  area  in  the  latter  case  is  3^  times  its  measure 
in  the  former  case.  And  as  the  same  relations  are  evidently 
true  for  any  whole  number  as  well  as  3, 

.'.  2.  If  any  segment  be  measured  by  an  assumed  u.l.  and 

also  by    th  of  the  assumed  u.l.  as  a  new  u.l..,  the  measure  of 
;/ 

the  segment  in  the  latter  case  is  n  times  its  measure  in  the 
former.     And  if  any  area  be  measured  by  the  corresponding 
u.a.%  the  measure  of  the  area  in  the  latter  case  is  n^  times  its 
measure  in  the  former  case  ;  n  being  any  whole  number. 
This  may  be  stated  otherwise  as  follows  : — 

By  reducing  an  assumed  u.l.  to  ^th  of  its  original  length, 

n 

we  increase  the  measure  of  any  given  segment  11  times,  and 
we  increase  the  measure  of  any  given  area  J1^  times  ;  ;/  being 
a  whole  number. 


104  SYNTHETIC   GEOMETRY. 

In  all  cases  where  a  u.l.  and  a  u.a.  are  considered  together, 
they  are  supposed  to  be  connected  by  the  relation  of  1 50°, 
3  and  4. 

I52^  Theorem. — The  number  of  unit-areas  in  a  rectangle 
is  the  product  of  the  numbers  of  unit-lengths  in  two  adjacent 
sides. 

The  proof  is  divided  mto  three  cases. 

1.  Let  the  measures  of  the  adjacent  sides  with  respect  to 
c    the  unit  adopted  be  whole  numbers. 

Let  AB  contain  the  assumed  11. l.  a 
times,  and  let  AD  contain  it  b  times. 
Then,  by  dividing  AB  into  a  equal 
parts  and  drawing,  through  each  point 
of  division,  lines  ||  to  AD,  and  by  dividing  AD  into  b  equal 
parts  and  drawing,  through  each  point  of  division,  lines  ||  to 
AB,  we  divide  the  whole  rectangle  into  equal  squares,  of 
which  there  are  a  rows  with  b  squares  in  each  row. 

the  whole  number  of  squares  is  ab. 
But  each  square  has  the  u.l.  as  its  side  and  is  therefore  the  u.a. 

u.a.s  in  AC=^?/./.s  in  AB  x  u.l.s  in  AD. 
We  express  this  relation  more  concisely  by  writing  symbolic- 
ally [=]AC  =  AB.AD, 

where [=:AC  means" the  number  ofu.a.s  in[=]AC,"and  AB  and 
AD  mean  respectively  "the  numbers  of  w./.s  in  these  sides." 

And  in  language  we  say,  the  area  of  a  rectangle  is  the  pro- 
duct of  its  adjacent  sides  ;  the  proper  interpretation  of  which 
is  easily  given. 

2.  Let  the  measures  of  the  adjacent  sides  with  respect  to 
the  unit  adopted  be  fractional. 

Then,  ".•  AB  and  AD  are  commensurable,  some  unit  will 
be  an  aliquot  part  of  each  (150°,  5).  Let  the  new  unit  be 
-th  of  the  adopted  unit,  and  let  AB  contain  p  of  the  new 

units,  and  AD  contain  ^  of  them. 

The  measure  of  [z:AC  in  terms  of  the  new  71. a.  is   ^a 


MEASUREMENT  OF  LENGTHS  AND  AREAS.    I05 

(i  52°,  i),  and  the  measure  of  the  ciiAC  in  terms  of  the  adopted 

unit  is  ^f  (151°,  2) 

But  the  measure  of  AB  in  terms  of  the  adopted  //./.  is 

'^,  and  of  AD  it  is  ^ .  (151",  2) 

n  n 

and  -^=^.2, 

or  nAC^AB.AD. 

Illus. — Suppose  the  measures  of  AB  and  AD  to  some 
unit-length  to  be  3.472  and  4.631.  By  taking  a  //./.  1000 
times  smaller  these  measures  become  the  whole  numbers 
3472  and  4631,  and  the  number  of  corresponding  u.a.s  in  the 
rectangle  is  3472  x  4631  or  16078832  ; 

and  dividing  by  looo^,  the  measure  of  the  area  with  respect  to 
the  original  2i.L  is     16.078832  =  3.472  x  4.631. 

3.  Let  the  adjacent  sides  be  incommensurable.     There  is 

now  no  ti.l.  that  will   measure  both  AB     b c 

and  AD. 

If  [=]AC  is  not  equal  to  AB .  AD,  let  it 


be  equal   to  AB  .  AE,  where  AE  has  a    a  e  h  d 

measure  different  from  AD  ;  and  suppose,  first,  that  AE  is  < 
AD,  so  that  E  lies  between  A  and  D. 

With  any  u.l.  which  will  measure  AB,  and  which  is  less 
than  ED,  divide  AD  into  parts.  One  point  of  division  at 
least  must  fall  between  E  and  D  ;  let  it  fall  at  H.  Complete 
the  rectangle  BH. 

Then  AB  and  AH  are  commensurable,  and 
□BH^AB.AH, 
but  miBD^AB.AE;  (hyp.) 

and  i=]BHis<[=!BD; 

AB.AH  is<AB.AE, 
and  AB  being  a  common  factor 

AH  is  <  AE  ;  which  is  not  true. 
.*.  If  [=iAC  =  AB  .  AE,  AE  cannot  be  <  AD,  and  similarly 
it  may  be  shown  that  AE  cannot  be  >  AD  ;  .*.  AE  =  AD,  or 
[=]AC=AB.AD.  q.e.d. 


i06  SYNTHETIC   GEOMETRY. 

153°.  The  results  of  the  last  article  in  conjunction  with 
Section  I.  of  this  Part  give  us  the  following  theorems. 

1.  The  area  of  a  parallelogram  is  the  product  of  its  base 
and  altitude.  (140°) 

2.  The  area  of  a  triangle  is  one-half  the  product  of  its  base 
and  altitude.  (141°) 

3.  The  area  of  a  trapezoid  is  one-half  the  product  of  its 
altitude  and  the  sum  of  its  parallel  sides.  (M5°j  Ex.) 

4.  The  area  of  any  regular  polygon  is  one-half  the  product 
of  its  apothem  and  perimeter.  (H7°) 

5.  The  area  of  a  circle  is  one-half  the  product  of  its  radius 
and  a  line-segment  equal  to  its  circumference.  (H9°) 

Ex.  I.  Let  O,  O'  be  the  centres  of  the  in-circle  and  of  the 

P  ex-circle  to  the  side  BC  (131°); 

V      y/^  \  and    let    OD,    O'P"  be   perpen- 

^^" \       diculars  on  BC,   OE,  O'P'  per- 

^(^;\d^^^^^^°'   pendiculars  on  AC,  and  OF,  O'P 
yy^^^^^^\y^/         ^"^  AB-     Then 

A  ^  c*\~P'    ^,^^1      0'P  =  0'P'  =  0'P''=r'; 

.-.  AABC  =  AAOB-|-ABOC  +  ACOA 

=  UB.  OF  +  ^BC.OD-t-|CA.OE  (153°,  2) 

=  ^  r  X  perimeter  =  7's, 
where  s  is  the  half  perimeter  ; 
/\  =  rs. 

Ex.  2.     AABC  =  AAO'B  +  AAO'C-ABO'C 

=  \0'V .  AB  -f-  ^O'P'.  AC  -  |0'P".  BC 

r=\r\b^c-a)  =  r\s-d), 
where  r'  is  the  radius  of  the  ex-circle  to  side  a  ; 
l\=r\s-a). 
Similarly,      A  =  ^"''■^  -  b) 


MEASUREMENT  OF  LENGTHS  AND  AREAS.    lO; 

Exercises. 

1.  l=L+J  +JL. 

I '    If  '    tit 
r    r     r      r 

2.  /\^  =  rr'r"r"'. 

3.  What  relation  holds  between  the  radius  of  the  in-circle 

and  that  of  an  ex- circle  when  the  triangle  is  equiangular.'* 
Note. — When  the  diameter  of  a  circle  is  taken  as  the  u.l. 
the  measure  of  the  circumference  is  the  inexpressible  numeri- 
cal quantity  symbolized  by  the  letter  tt,  and  which,  expressed 
approximately,  is  3. 141 5926.... 

4.  What  is  the  area  of  a  square  when  its  diagonal  is  taken 

as  the  u.l.  ? 

5.  What  is  the  measure  of  the  diagonal  of  a  square  when 

the  side  is  taken  as  the  ?/./.?  (150°,  5) 

6.  Find  the  measure  of  the  area  of  a  circle  when  the  di- 

ameter is  the  u.l.     When  the  circumference  is  the  u.l. 

7.  If  one  line-segment  be  twice  as  long  as  another,  the 

square  on  the  first  has  four  times  the  area  of  the 
square  on  the  second.  (151°)  2) 

8.  If  one  line-segment  be  twice  as  long  as  another,  the 

equilateral  triangle  on  the  first  is  four  times  that  on 
the  second.  (141°) 

9.  The  equilateral  triangle  on  the  altitude  of  another  equilat- 

eral triangle  has  an  area  three-fourths  that  of  the  other. 

10.  The  three  medians  of  any  triangle  divide  its  area  into 

six  equal  triangles. 

1 1.  From  the  centroid  of  a  triangle  draw  three  lines  to  the 

sides  so  as  to  divide  the  triangle  into  three  equal 
quadrangles. 

12.  In  the  triangle  ABC  X  is  taken  in  BC,  Y  in  CA,  and  Z 

in  AB,  so  that  BX  =  iBC,  CY  =  iCA,  and  AZ  =  ^AB. 
Express  the  area  of  the  triangle  XYZ  in  terms  of  that 
of  ABC. 

13.  Generalize  12  by  making  BX  =  -BC,  etc. 


14.  Show  that  a  =  s(\-  ^\  =  -{r" ■\-r' 


108  SYNTHETIC  GEOMETRY. 


SECTION    III. 

GEOMETRIC  INTERPRETATION  OF 
ALGEBRAIC  FORMS. 

1 54''.  We  have  a  language  of  symbols  by  which  to  express 
and  develop  mathematical  relations,  namely,  Algebra.  The 
symbols  of  Algebra  are  quantitative  and  operative,  and  it  is 
very  desirable,  while  giving  a  geometric  meaning  to  the 
symbol  of  quantity,  to  so  modify  the  meanings  of  the  sym- 
bols of  operation  as  to  apply  algebraic  forms  in  Geometry. 
This  application  shortens  and  generalizes  the  statements  of 
geometric  relations  without  interfering  with  their  accuracy. 

Elementary  Algebra  being  generalized  Arithmetic,  its 
quantitative  symbols  denote  numbers  and  its  operative  sym- 
bols are  so  defined  as  to  be  consistent  with  the  common 
properties  of  numbers. 

Thus,  because  2  +  3  =  3-1-2  and  2.3  =  3.2,  we  say  that 
a^rb  =  b-\-a  and  ab=ba. 

This  is  called  the  commutative  law.  The  first  example  is 
of  the  existence  of  the  law  in  addition,  and  the  second  of  its 
existence  in  multiplication. 

The  commutative  law  in  addition  may  be  thus  expressed : — 
A  sum  is  independent  of  the  order  of  its  addends  ;  and  in 
multiplication — A  product  is  independent  of  the  order  of  its 
factors. 

Again,  because  2(3  +  4)  =  2  .  3  -f-  2  .  4,  we  say  that 
a{b  +  c)  —  ab  +  ac. 

This  is  called  the  distributive  law  and  may  be  stated 
thus  : — The  product  of  multiplying  a  factor  by  the  sum  of 
several  terms  is  equal  to  the  sum  of  the  products  arising  from 
multiplying  the  factor  by  each  of  the  terms. 

These  two  are  the  only  laws  which  need  be  here  mentioned. 
And  any  science  which  is  to  employ  the  forms  of  Algebra 


INTERPRETATION  OF  ALGKBRAIC  FORMS.  IO9 

must  have  that,  whatever  it  may  be,  which  is  denoted  by  the 
algebraic  symbol  of  quantity,  subject  to  these  laws. 

155°.  As  already  explained  in  22°  we  denote  a  single  hne- 
segment,  in  the  one-letter  notation,  by  a  single  letter,  as  a, 
which  is  equivalent  to  the  algebraic  symbol  of  quantity  ;  and 
hence, 

A  single  algebraic  symbol  of  quantity  is  to  be  interpreted 
geometrically  as  a  line- segment. 

It  must  of  course  be  understood,  in  all  cases,  that  in  em- 
ploying the  two-letter  notation  for  a  segment  (22°),  as  "  AB," 
the  two  letters  standing  for  a  single  line-segment  are  equiva- 
lent to  but  a  single  algebraic  symbol  of  quantity. 

The  expression  a^b  denotes  a  segment  equal  in  length  to 
those  denoted  by  a  and  b  together. 

Similarly  '2a  =  a-\-a^  and  na  means  a  segment  as  long  as  n 
of  the  segments  a  placed  together  in  line,  n  being  any 
numerical  quantity  whatever.  (28°) 

a  —  b^  when  a  is  longer  than  b,  is  the  segment  which  is  left 
when  a  segment  equal  to  b  is  taken  from  a. 

Now  it  is  manifest  that,  if  a  and  b  denote  two  segments, 
a-vb=b-\-a,  and  hence  that  the  commutative  law  for  addition 
applies  to  these  symbols  when  they  denote  magnitudes  having 
length  only,  as  well  as  when  they  denote  numbers. 

1 56°.  Line  in  Opposite  Senses. — A  quantitative  symbol,  a,  is 
in  Algebra  always  affected  with  one  of  two  signs,  -f  or  - , 
which,  while  leaving  the  absolute  value  of  the  symbol  un- 
changed, impart  to  it  certain  properties  exactly  opposite  in 
character. 

This  oppositeness  of  character  finds  its  complete  interpreta- 
tion in  Geometry  in  the  opposite  directions  of  every  segment. 

Thus  the  segment  in  the  margin  may  be  con-  a 

sidered  as  extending/r<77/z  A  to  B  or  from  B  to  A.  A  B 

With  the  two-letter  notation  the  direction  can  be  denoted 
by  the  order  of  the  letters,  and  this  is  one  of  the  advantages 


I  TO  SYNTHETIC   GEOMETRV. 

of  this  notation ;  but  with  the  one-letter  notation,  if  we  denote 
the  segment  AB  by  +  a^  we  ;;/«j-/  denote  the  segment  B A 
by  -a. 

But  as  there  is  no  absolute  reason  why  one  direction  rather 
than  the  other  should  be  considered  positive,  we  express  the 
matter  by  saying  that  AB  and  BA,  or  +a  and  -a,  denote 
the  same  segment  taken  in  opposite  senses. 

Hence  the  algebraic  distinction  of  positive  and  negative  as 
applied  to  a  single  symbol  of  quantity  is  to  be  interpreted 
geometrically  by  the  oppositeness  of  direction  of  the  segment 
denoted  by  the  symbol. 

Usually  the  applications  of  this  principle  in  Geometry  are 
confined  to  those  cases  in  which  the  segments  compared  as 
to  sign  are  parts  of  one  and  the  same  line  or  are  parallel. 

Ex.  I.  Let  ABC  be  any  A  and  let  BD  be  the  altitude  from 
the  vertex  B. 

Now,  suppose  that  the  sides  AB  and  BC 
undergo  a  gradual  change,  so  that  B  may 
move  along  the  Hne  BB'  until  it  comes 
into  the  position  denoted  by  B'. 
"a  d  c  Then  the  segment  AD  gradually  di- 
minishes as  D  approaches  A  ;  disappears  when  D  coincides 
with  A,  in  which  case  B  comes  to  be  vertically  over  A 
and  the  A  becomes  right-angled  at  A ;  reappears  as  D 
passes  to  the  left  of  A,  until  finally  we  may  suppose  that 
one  stage  of  the  change  is  represented  by  the  AAB'C  with 
its  altitude  B'D'. 

Then,  if  we  call  AD  positive,  we  must  call  AD'  negative,  or 
we  must  consider  AD  and  AD'  as  having  opposite  senses. 

Again,  from  the  principle  of  continuity  (104°)  the  foot  of 
the  altitude  cannot  pass  from  D  on  the  right  of  A  to  D'  on 
the  left  of  A  without  passing  through  every  intermediate 
point,  and  therefore  passing  through  A.  And  thus  the  seg- 
ment AD  must  vanish  before  it  changes  sign. 

This   is   conveniently  expressed   by   saying  that  a   line- 


INTERPRETATION  OF  ALGEBRAIC  FORMS.     1 1 1 

segjnent  changes  sign  when  it  passes  through  zero;  passing 
through  zero  being  interpreted  as  vanishing  and  reappearing 
on  the  other  side  of  the  zero-point. 

Ex.  2.  ABCD  is  a  normal  quadrangle.     Consider  the  side 
AD  and  suppose  D  to  move  along  the  line         g 
DA  until  it  comes  into  the  position  D'. 

The  segments  AD  and  AD'  are  opposite 
in  sense,  and  ABCD'  is  a  crossed  quad- 
rangle. 

.•.  the  crossed  quadrangle  is  derived  from       "^       ^  ° 

the  normal  one  by  changing  the  sense  of  one  of  the  sides. 

Similarly,  if  one  of  the  sides  of  a  crossed  quadrangle  be 
changed  in  sense  the  figure  ceases  to  be  a  crossed  quadrangle. 

Ex.  3.  This  is  an  example  where  segments  which  are  par- 
allel but  which  are  not  in  line  have 
opposite  senses. 

ABC  is  a  A  ^rid  P  is  any  point 
within  from  which  perpendiculars 
PD,  PE,  PF  are  drawn  to  the  sides. 

Suppose  that  P  moves  to  P'. 
Then  PF  becomes  P'F',  and  PF 
and  P'F'  being  in  the  same  direc- 
tion have  the  same  sense.  Similarly  PE  becomes  P'E', 
and  these  segments  have  the  same  sense.  But  PD  becomes 
P'D'  which  is  read  in  a  direction  opposite  to  that  of  PD. 
Hence  PD  and  P'D'  are  opposite  in  sense. 

But  PD  and  P'D'  are  perpendiculars  to  the  same  line  from 
points  upon  opposite  sides  of  it,  and  it  is  readily  seen  that  in 
passing  from  P  to  P'  the  _LPD  becomes  zero  and  then  changes 
sense  as  P  crosses  the  side  BC. 

Hence  if  by  any  continuous  change  in  a  figure  a  point 
passes  from  one  side  of  a  line  to  the  other  side,  the  peipen- 
dicular  from  that  point  to  the  line  changes  sense. 

Cor,   If  ABC  be  equilateral  it  is  easily  shown  that 
PD-}-PE-hPF  =  a  constant. 


112  SYNTHETIC   GEOMETRY. 

And  if  we  regard  the  sense  of  the  segments  this  statement  is 
true  for  all  positions  of  P  in  the  plane. 

157°.  Product. — The  algebraic  form  of  a  product  of  two 
symbols  of  quantity  is  interpreted  geometrically  by  the  rect- 
angle having  for  adjacent  sides  the  segments  denoted  by  the 
quantitative  symbols. 

This  is  manifest  from  Art.  152°,  for  in  the  form  ab  the 
single  letters  may  stand  for  the  measures  of  the  sides,  and  the 
product  ab  will  then  be  the  measure  of  the  area  of  the  rect- 
angle. 

If  we  consider  ab  as  denoting  a  en  having  a  as  altitude  and 
b  as  base,  then  ba  will  denote  the  izu  having  b  as  altitude  and 
a  as  base.  But  in  any  en  it  is  immaterial  which  side  is  taken 
as  base  (138°)  ;  therefore  ab^ba,  and  the  form  satisfies  the 
commutative  law  for  multiplication. 

Again,  let  AC  be  the  segment  b-\-Cy  and  AB  be  the  segment 

A  D c   a,  so  placed  as  to  form  the  ni^a{b  +  c)  or 

"  ^         I    AF.     Taking  AD  =^^,  let  DE  be  drawn 

!    II  to  AB.     Then  AE  and  DF  are  rect- 
F    angles  and  DE  =  AB  =  rt:. 
cziAE  is  nnaby  and  cziDF  is  nnac  ; 
n2a{b  ->rc)  =  nnab  -f  \:nac^ 
and  the  distributive  law  is  satisfied. 

158°.  We  have  then  the  two  following  interpretations  to 
which  the  laws  of  operation  of  numbers  apply  whenever  such 
operations  are  interpretable. 

I.  A  si7tgle  symbol  of  qua7itity  denotes  a  line -segment. 

As  the  sum  or  difference  of  two  line-segments  is  a  segment, 
the  sum  of  any  number  of  segments  taken  in  either  sense  is  a 
segment. 

Therefore  any  number  of  single  symbols  of  quantity  con- 
nected by  +  and  -  signs  denotes  a  segment,  as  a-\-b^ 
a-b-irc,  a-b  +  {-c),  etc. 


INTERPRETATION  OF  ALGEBRAIC  FORMS.     II 3 

For  this  reason  such  expressions  or  forms  are  often  called 
linear^  even  in  Algebra. 

Other  forms  of  linear  expressions  will  appear  hereafter. 

2.  The  product  form  of  two  symbols  of  quantity  denotes  the 
rectangle  whose  adjaceitt  sides  are  the  segments  denoted  by  the 
single  symbols. 

A  rectangle  encloses  a  portion  of  the  plane  and  admits  of 
measures  in  two  directions  perpendicular  to  one  another, 
hence  the  area  of  a  rectangle  is  said  to  be  of  two  dimensions. 
And  as  all  areas  can  be  expressed  as  rectangles,  areas  in 
general  are  of  two  dimensions. 

Hence  algebraic  terms  which  denote  rectangles,  such  as  ab^ 
{a-\-b)c,  {a-irb){c+d\  etc.,  are  often  called  rectangular  terms, 
and  are  said  to  be  of  two  dimensions.        a  d     c 


Ex.  Take  the  algebraic  identity 

a{b-\-c)  =  ab-\-ac. 

The  geometric  interpretation  gives —     be  f 

If  there  be  any  three  segments  {a,  b,  c)  the  cu  on  the  first 
and  the  sum  of  the  other  two  {b,  c)  is  equal  to  the  sum  of  the 
czis  on  the  first  and  each  of  the  other  two. 

The  truth  of  this  geometric  theorem  is  evident  from  an 
inspection  of  a  proper  figure. 

This  is  substantially  Euclid,  Book  II.,  Prop.  i. 


159°.  Square. — When  the  segment  b  is  equal  to  the  seg- 
ment a  the  rectangle  becomes  the  square  on  a.  When  this 
equality  of  symbols  takes  place  in  Algebra  we  write  a^  for  aa, 
and  we  call  the  result  the  "  square  "  of  a,  the  term  "  square  " 
being  derived  from  Geometry. 

Hence  the  algebraic  form  of  a  square  is  interpreted  geo- 
metrically by  the  square  which  has  for  its  side  the  segment 
denoted  by  the  root  symbol. 

Ex.  In  the  preceding  example  let  b  become  equal  to  a^  and 
a{a  +  ^)  =  ^-  4-  iiCj 
H 


114  SYNTHETIC   GEOMETRY. 

which  interpreted  geometrically  gives — 

If  a  segment  (a  +  c)  be  divided  into  two  parts  {a,  c),  the 
rectangle  on  the  segment  and  one  of  its  parts  {a)  is  equal  to 
the  sum  of  the  square  on  that  part  {aP)  and  the  rectangle  on 
the  two  parts  {ac). 

This  is  Euclid,  Book  II.,  Prop.  3.  The  truth  of  the  geo- 
metric theorem  is  manifest  from  a  proper  figure. 

160°.  Homogeneity. — Let  a,  b,  c,  d  denote  segments.  In 
the  linear  expressions  a  +  b,  a-b,  etc.,  and  in  the  rectangular 
expressions  ab  +  cd,  etc.,  the  interpretations  of  the  symbols  + 
and  —  are  given  in  28°,  29°,  and  143°,  and  are  readily  in- 
telligible. 

But  in  an  expression  such  as  ab  +  c  we  have  no  interpreta- 
tion for  the  symbol  -f-  if  the  quantitative  symbols  denote 
line-segments.  For  ah  denotes  the  area  of  a  rectangle  and  c 
denotes  a  segment,  and  the  adding  of  these  is  not  intelligible 
in  any  sense  in  which  we  use  the  word  "  add." 

Hence  an  expression  such  as  ab  +  c  is  not  interpretable 
geometrically  This  is  expressed  by  saying  that — An  alge- 
braic form  has  no  geometric  interpretation  unless  the  form  is 
homogeneous,  i.e.,  unless  each  of  its  terms  denotes  a  geo- 
metric element  of  the  same  kind. 

It  will  be  observed  that  the  terms  "square,"  "dimensions," 
"homogeneous,"  and  some  others  have  been  introduced  into 
Algebra  from  Geometry. 

161°.  Rectangles  i7i  Opposite  Senses. — The  algebraic  term 
ab  changes  sign  if  one  of  its  factors  changes  sign.  And  to  be 
consistent  we  must  hold  that  a  rectangle  changes  sense 
whenever  one  of  its  adjacent  sides  changes  sense. 

Thus  the  rectangles  AB .  CD  and  AB .  DC  are  the  same  in 
extent  of  area,  but  have  opposite  senses.     And 

AB.CD  +  AB.DC  =  o, 
for  the  sum  =  AB(CD  +  DC), 

and  CD  +  DC  =  o.  (156°) 


INTERPRETATION  OF  ALGEBRAIC  FORMS. 


15 


+ 

— 

0 

-f 

c 


As  the  sense  of  a  rectangle  depends  upon  that  of  a  line- 
segment  there  is  no  difficulty  in  determining  when  rectangles 
are  to  be  taken  in  different  senses. 

The  following  will  illustrate  this  part  of  the  subject  : — 

Let  OA  =  OA'and  OC  =  OC',  and  let     bob, 
the  figures  be  rectangles. 

ns  OA.  OC  and  OA'.  OC  have  the 
common  altitude  OC  and  bases  equal 
in  length  but  opposite  in  sense.    There- 
fore OA.'OC  and  OA'.  OC  are  opposite  in  sense,  and  if  we 
call  [zdOA.  OC  positive  we  must  call  cdOA'.  OC  negative. 

Again,  cds  OC  .  OA'  and  OC .  OA'  have  the  common  base 
OA'  and  altitudes  equal  in  length  but  opposite  in  sense. 
Therefore  cds  OC.  OA'  and  OC.  OA'  are  opposite  in  sense, 
and  therefore  ens  OA.  OC  and  OA'.  OC  are  of  the  same  sense. 
Similarly  ens  OC .  OA'  and  OC.  OA  are  of  the  same  sense. 

These  four  czis  are  equivalent  to  the  algebraic  forms  : — 
+  a.  +  l>=  +a^,  -a.-\-b=  -ab, 

-{■a.-b=  -ab,  -a.-b=+ab. 

Ex.  I.  ABCD  is  a  normal  quadrangle  whose  opposite  sides 
meet  in  O,  and  OE,  OF  are  altitudes 
of  the  As  DOC  and  AOB  respectively. 
The  Od.  ABCD 

=ADOC-AAOB, 
=  |c=]DC.OE-|[=]AB.OF.  (141°) 
Now,  let  A  move   along  AB   to  A' 
(104°).     Then  O  comes  to  O',  F  to  F', 
E  to  E',  and  O'E',  O'F'  become  the  altitudes  of  the  As 
DO'C  and  A'O'B  respectively. 

But  O'E'  and  OE  have  the  same  sense,  therefore  DC .  OE 
and  DC .  OE'  have  the  same  sense. 

Also,  A'B  is  opposite  in  sense  to  AB,  and  O'F'  is  opposite 
in  sense  to  OF.  (156°,  Ex.  3) 

AB  .  OF  and  A'B  .  O'V  have  the  same  sense  ; 
Od.  A'BCD  =  ADO'C-AA'0'B; 


ii6 


SYNTHETIC   GEOMETRY. 


/  - 

yr 

N 

/\ 

\U 

^ 

J 

s 

D 

R 

.^Q 


or,  the  area  of  a  crossed  quadrangle  must  be  taken  to  be  the 
difference  between  the  two  triangles  which  constitute  it. 

162°.  Theorem. — A  quadrangle  is  equal  to  one-half  the 
parallelogram  on  its  diagonals  taken  in  both  magnitude  and 
relative  direction. 

P  B  .Q  ABCD  is  a  quadrangle  of  which  AC 
and  BD  are  diagonals.  Through  B  and 
D  let  PQ  and  RS  be  drawn  ||  to  AC,  and 
through  A  and  C  let  PS  and  OR  be 
drawn  ||  to  BD.  Then  PORS  is  the 
I  7  on  the  diagonals  AC  and  BD  in 
both  magnitude  and  direction. 
Qd.  ABCD  =  |z=::7PQRS. 

Proof.— Qi^.  ABCD 

=AABC  +  AADC(istFig.) 
= AABC  -  AADC  (2ndFig.) 
A  (161°,  Ex.) 

AABC  =  i£II7P0CA, 

AADC  =  |nZ7SRCA,  (141°,  Cor.  i) 

Qd.  ABCD=^£ZI7P0RS  in  both  figures. 
This  theorem  illustrates  the  generality  of  geometric  results 
when  the  principle  of  continuity  is  observed,  and  segments 
and  rectangles  are  considered  with  regard  to  sense.  Thus 
the  principle  of  continuity  shows  that  the  crossed  quadrangle 
is  derived  from  the  normal  one  (156°,  Ex.  2)  by  changing  the 
sense  of  one  of  the  sides. 

This  requires  us  to  give  a  certain  interpretation  to  the  area 
of  a  crossed  quadrangle  (161°,  Ex.  i),  and  thence  the  present 
example  shows  us  that  all  quadrangles  admit  of  a  common 
expression  for  their  areas. 


163°.  A  rectangle  is  constructed  upon  two  segments  which 
are  independent  of  one  another  in  both  length  and  sense. 
But  a  square  is  constructed  upon  a  single  segment,  by  using 


INTERPRF.TATION  OF  ALGEBRAIC  FORMS.  11/ 

it  for  each  side.  In  other  words,  a  rectangle  depends  upon 
two  segments  while  a  square  depends  upon  only  one. 

Hence  a  square  can  have  only  one  sign,  and  this  is  the 
one  which  we  agree  to  call  positive. 

Hence  a  squtwe  is  always  positive. 

164°.  The  algebraic  equation  ab=cd  tells  us  geometrically 
that  the  rectangle  on  the  segments  a  and  b  is  equal  to  the 
rectangle  on  the  segments  c  and  d. 

But  the  same  relation  is  expressed  algebraically  by  the  form 
cd 

therefore,  since  rz  is  a  segment,  the  form  ^  is  linear  and 

o 

denotes  that  segment  which  with  a  determines  a  rectangle 
equal  to  cd. 

Hence  an  expression  such  as     -  +  —  +  —  is  linear. 

cab 

165°.  The  expression  a^  =  bc  tells  us  geometrically  that  the 
square  whose  side  is  a  is  equal  to  the  rectangle  on  the  seg- 
ments b  and  c. 

But  this  may  be  changed  to  the  form 
a  =  J  be. 
Therefore  since  «:  is  a  segment,  the  side  of  the  square,  the  form 
f^fbc  is  linear. 

Hence  the  algebraic  forin  of  the  square  root  of  the  product 
of  two  symbols  of  quantity  is  interpreted  geometrically  by  the 
side  of  the  square  which  is  equal  to  the  rectangle  on  the 
segments  denoted  by  the  quatititative  symbols. 

166°.  The  following  theorems  are  but  geometric  interpreta- 
tions of  well-known  algebraic  identities.  They  may,  however, 
be  all  proved  most  readily  by  superposition  of  areas,  and 
thus  the  algebraic  identity  may  be  derived  from  the  geo- 
metric theorem. 


Ii8 


SYNTHETIC  GEOMETRY. 


ab 

b' 

a« 

ab 

I.  The  square  on  the  sum  of  two  segments  is  equal  to 
the   sum    of  the   squares   on   the  segments  and   twice  the 
a  b      rectangle  on  the  segments. 

2.  The  rectangle  on  the  sum  and  differ- 
ence of  two  segments  is  equal  to  the 
difference  of  the  squares  on  these  segments. 

3.  The  sum  of"  the  squares  on  the  sum  and  on  the  differ- 
ence of  two  segments  is  equal  to  twice  the  sum  of  the  squares 
on  the  segments. 

{a  +  bf  +  {a-  bf  =  2(^2  +  d'^)^  a>b. 

4.  The  difference  of  the  squares  on  the  sum  and  on  the 
difference  of  two  segments  is  equal  to  four  times  the  rectangle 
on  the  segments. 

{a^-bf-{a-bf  =  ^ab,  a>b. 

Exercises. 

I.  To  prove  4  of  Art.  166°. 

Let  AH  =  ^  and  HB  =  ^  be  the  segments,  so  that 
A  g         H  6   B    AB   is   their  sum.     Through    H 

draw  HG  |1  to  EC,  a  side  of  the 
square  on  AB.  Make  HG  =  rt, 
and  complete  the  square  FGLE, 
as  in  the  figure,  so  that  FG  is 
a  —  b. 

Then  AC  is  {a  +  bf  and  EG  is 
{a  — by-,    and  their  difference  is 
the  four  rectangles  AF,  HK,  CL,  and  DE  ;  but  these 
each  have  a  and  b  as  adjacent  sides. 
{a  +  bf-{a-b)'^=^ab. 
State  and  prove  geometrically  {a  -  b)^= or  +  b^  -  2ab. 
State  and  prove  geometrically 

{a-\-b){a-{-c)  =  a'-  +  a{b  +  c)  +  bc. 
State  and  prove  geometrically  by  superposition  of  areas 


ab          b 

ab 
b 

b 
ab 

E             F 
L            G 

6          ab 

(«+b)'^ 


AREAL    RELATIONS.  1  I9 

{a-\-i?)--\-{a-bf-\-2{(i  +  b){a-b)=^{2ay,v^-hQVQ  a  and  b 

denote  segments. 
If  a  given  segment  be  divided  into  any  three  parts  the 

square  on  the  segment  is  equal  to  the  sum  of  the 

squares  on  the  parts  together  with  twice  the  sum  of 

the  rectangles  on  the  parts  taken  two  and  two. 
Prove,  by  comparison  of  areas  from  the  Fig.  of  Ex.  i,  that 

{a-\-bf  =  2b{a  +  b)  +  2b{a'-b)->r{a-bf,    and    state    the 

theorem  in  words. 


SECTION    IV. 


AREAL  RELATIONS. 

167°.  Def. —  1.  The  segment  which  joins  two  given  points 
is  called  the  join  of  the  points  ;  and  where  no  reference  is 
made  to  length  the  Join  of  two  points  may  be  taken  to  mean 
the  hne  determined  by  the  points. 

2.  The  foot  of  the  perpendicular  from  a  given  point  to  a 
given  line  is  the  orthogonal  projection^  or  simply  the  projec- 
tion^ of  the  point  upon  the  line. 

3.  Length  being  considered,  the  join  of  the  projection  of 
two  points  is  the  projection  of  the 
join  of  the  points. 

Thus  if  L  be  a  given  line  and  P, 
Q,  two  given  points,  and  PP',  QQ' 
perpendiculars  upon  L  ;  PQ  is  the 
join  of  P  and  Q,  P'  and  Q'  are  the  p^  Q^      l 

projections  of  P  and  Q  upon  L,  and  the  segment  P'Q'  is  the 
projection  of  PO  upon  L. 

168°.   Theorem.— T\iQ:  sum  of  the  projections  of  the  sides  of 


20 


SYxNTHETIC    GEOMETRY. 


any   closed    rectilinear   figure,    taken   in   cyclic   order  with 
c  respect  to  any  line,  is  zero. 

ABCD    is    a    closed    rectilinear 
figure  and  L  is  any  line.     Then 
Pr.AB  +  Pr.BC  +  Pr.CD  +  Pr.DA=o 

A'    B  o       D'   L       /'/T?^— Draw  the  perpendiculars 

AA',  BE',  CC,  DD',  and  the  sum  of  the  projections  becomes 

A'B'  +  B'C'  +  C'D'  +  D'A'. 
But  D'A'  is  equal  in  length  to  the  sum  of  the  three  others  and 
is  opposite  in  sense.         .*.  the  sum  is  zero. 

It  is  readily  seen  that  since  we  return  in  every  case  to  the 
point  from  which  we  start  the  theorem  is  true  whatever  be 
the  number  or  disposition  of  the  sides. 

This  theorem  is  of  great  importance  in  many  investigations. 

Cor.  Any  side  of  a  closed  rectilinear  figure  is  equal  to  the 
sum  of  the  projections  of  the  remaining  sides,  taken  in  cyclic 
order,  upon  the  line  of  that  side. 

Def. — In  a  right-angled  triangle  the  side  opposite  the  right 
angle  is  called  the  hypothe?tuse,  as  distinguished  from  the 
remaining  two  sides. 

Theorem. — In  any  right-angled  triangle  the  square 
on  one  of  the  sides  is  equal  to  the  rectangle 
on  the  hypothenuse  and  the  projection  of 
that  side  on  the  hypothenuse. 

ABC  is  right-angled  at  B,  and  BD  is  ± 
AC.     Then  AB2  =  AC.AD. 

Proof.— \.^\.  AF  be  the  □  on  AC,  and  let 
EH  be  II  to  AB,  and  AGHB  be  a  en,  since 
Z_B  is  a  ~|- 

Then-.-  ^GAB  =  z.EAC=~|,  (82^  Cor.  5) 

z.CAB  =  ^EAG. 
Also,  AE  =  AC,  (hyp.) 

ACAB  =  AEAG,  (64°) 

and        .-.  AG  =  AB,  and  AH  is  the  n  P" '^R- 


AREAL    RELATIONS. 


121 


Now  nAH=-£IZ7ABLE  =  i=]ADKE,      (140°) 

i.e.,  AB'-^  =  AC.AD.  q.e.d. 

As  this  theorem  is  very  important  we  give  an  alternative 
proof  of  it. 

Proof.— KY  is  the  D  on  AC  and  AH 
is  the  D  on  AB,  and  BD  is  _L  AC. 

^GAB=Z.CAE=~1.        (82°,  Cor.  5)G< 


.*. 

^GAC=z.BAE. 

Also, 

AG  =  AB, 

and 

AC  =  AE, 

(82°,  Cor^  5) 

.'. 

AGAC=ABAE. 

(5O 

But 

AGAC  =  inAH, 

(141°) 

and 

ABAE=|i=iAK, 
DAH=dAK, 

i.e.. 

AB2=AC.AD. 

Cor.  I.  Since  AB2  =  AC  .  AD  we  have  from  symmetry 
BC2=AC.DC, 
.-.  adding,  AB2  +  BC2  =  AC(AD  +  UC), 

or  AB2  +  BC2  =  AC2. 

.'.  The  square  on  the  hypothenuse  of  a  right-angled  triangle 
is  equal  to  the  sum  of  the  squai'es  on  the  remaining  sides. 

This  theorem,  which  is  one  of  the  most  important  in  the 
whole  of  Geometry,  is  said  to  have  been  discovered  by 
Pythagoras  about  540  B.C. 

Cor.  2.  Denote  the  sides  by  a  and  c  and  the  hypothenuse 
by  b,  and  let  a-^  and  c^  denote  the  projections  of  the  sides  a 
and  c  upon  the  hypothenuse. 

Then  a^  =  af),    c'^^c^b., 

and  «2  +  ^2_^2_ 

Cor.  3.  Denote  the  altitude  to  the  hypothenuse  by  p. 
Then  b=^c^-\-a^,  and  ADB  and  CDB  are  right-angled  at  D, 
^2= ^^2  + ^^2 .,.2^^^^  ;  (166°,  i) 

add  2/J2  to  each  side  and 

b^  +  ip^  =  c;^  -1-/2  +  a^  +/2  +  2r^.7„ 


122  SYNTHETIC   GEOiMETRY. 

or  6-^  +  rt2_|.2^2_^2  +  ^2  +  2^i^i.  (Cor.  l) 

or  BD2  =  AD.DC, 

i.e.^  the  square  on  the  altitude  to  the  hypothenuse  is  equal  to 
the  rectangle  on  the  projections  of  the  sides  on  the  hypo- 
thenuse. 

Def. — The  side  of  the  square  equal  in  area  to  a  given 
rectangle  is  called  the  mean  proportional  or  the  geoinetric 
mean  between  the  sides  of  the  rectangle. 

Thus  the  altitude  to  the  hypothenuse  of  a  right-angled  /\ 
is  a  geometric  mean  between  the  segments  into  which  the 
altitude  divides  the  hypothenuse.  (169°,  Cor.  3) 

And  any  side  of  the  A  is  a  geometric  mean  between  the 
hypothenuse  and  its  projection  on  the  hypothenuse.       (169'') 

170°.   Theorem. — If  the  square  on  one  side  of  a  triangle  is 
equal  to  the  sum  of  the  squares  on  the  remaining  sides,  the 
triangle  is  right-angled  at  that  vertex  which  is  opposite  the 
side  having  the  greatest  square.     (Converse  of  169°,  Cor.) 
If  AC2  =  AB2  +  BC^,  the  Z.B  is  a  "1- 

Proof.— l.tt  ADC  be  a  i0  on  AC. 
AC=^  =  AB2-fBC2, 
AB  is  <  AC. 
a  chord  AD  can  be  found  equal  to  AB. 
//  Then  the  AADC  is  right-angled  at  D. 

IS  (106",  Cor.  4) 

AC-  =  AD2-fDC2,  (169°,  Cor.  i) 

and  AC2=AB2-fBC2,  and  AD  =  AB.  (hyp.) 

DC  =  BC, 
and  AADC  =  AABC. 

^D=^B  =  ~1.  ^.^.^• 

171°.  Theorem  169°  with  its  corollaries  and  theorem  170° 
are  extensively  employed  in  the  practical  applications  of 
Geometry.     If  we  take  the  three  numbers  3,  4,  and  5,  we 


AREAL   RELATIONS. 


123 


have  52=32  +  42.  Therefore  if  a  triangle  has  its  sides  3,  4, 
and  5  feet,  metres,  miles,  or  any  other  ;/./.,  it  is  right-angled 
opposite  the  side  5. 

For  the   segments   into   which    the    altitude   divides  the 
hypothenuse  we  have  S^h^3'^  ^"^  5'^i  =  4^  whence  ^?i  =  |  and 


=iJi 


For  the  altitude  itself, /2==u 


whence  p  ■■ 


Problem. — To  find  sets  of  whole  numbers  which  represent 
the  sides  of  right-angled  triangles. 

This  problem  is  solved  by  any  three  numbers  ,r,  y^  and  s^ 
which  satisfy  the  condition  ;ir2  =^2  4.  ^2_ 

Let  in  and  n  denote  any  two  numbers.     Then,  since 

{7n^-Vn^Y  =  {in'^-n^f^{2inny,  (166°,  4) 

the  problem  will  be  satisfied  by  the  numbers  denoted  by 
)n^-\-n^^  m^  —  fi^,  and  2tfin. 

The   accompanying  table,   which    may   be    extended    at 
pleasure,  gives  a  number  of  sets  of  such  numbers  : — 


2 

3 

4 

5 

6 

7 

8 

9 

10  '  ... 

5  '  10 

15 

26 

37 

50 

65 

82 

lOI   ... 

I 

3    6 

8 

10 

12 

14 

16 

18 

20 

,, 

!  4  :  8 

17 

24 

3S 

48 

63 

80 

99  1 

,^3 

20 

29 

40 

53 

68 

85 

104 

2 

i  12 

16 

20 

24 

28 

32 

36 

40 

.. 

5 

12 

21 

32 

45 

60 

77 

96 

.. 

25 

34 

45 

S8 

7?> 

90 

109 

.. 

3  ' 

24 

30 

36 

42 

48 

54 

60 

.. 

' 

7 

16 

27 

40 

55 

72 

91 

.. 

41 

52 

6^ 

80 

97 

116 

.. 

4 

9 

48 

S6 

64 

72 

80 

.. 

40 

20 

33 

48 

65 

84 

61 

74 

89 

106 

125  J 

5 

60 

70 

80 

90 

icoj 

i    1 

" 

24 

39 

56 

75  1 

.. 

... 

... 

•• 

24  SYNTHETIC   GEOMETRY. 

172°.  Let  a,  d,  c  be  the  sides  of  any  triangle,  and  let  b  be 
3  taken  as  base.     Denote  the  projections 

of  a  and  r  on  /^  by  a-^  and  q,  and  the 
^a        altitude  to  b  by  p.     Then 

(i)     U^^c-i^-\-a^^2.c^a^,        (166°,  i) 

(2)  ^2  =  ^1^/2,  (169°,  Cor.  I) 

(3)  ^^  =  ^i2+/2. 

1.  By  subtracting  (2)  from  (3) 

: .  The  difference  between  the  squares  upon  two  sides  of  a 
triangle  is  equal  to  the  difference  of  the  squares  on  the  projec- 
tions of  these  sides  on  the  third  side,  taken  in  the  same  order. 

Since  all  the  terms  are  squares  and  cannot  change  sign 
(163°),  the  theorem  is  true  without  any  variation  for  all  As. 

2.  By  adding  (i)  and  (2)  and  subtracting  (3), 

b'^-\-c'^-a^  =  ic^  +  2^1  ^1 

=  2^^i,  •.*  b=c^-\-a^, 
a'^  =  b''-  +  c'^-2bc^. 
Now,  since  we  have  assumed  that  b=c-^^-]-a-^,  where  c^  and  a^ 
are  both  positive,  D  falls  between  A  and  C,  and  the  angle  A 
is  acute. 

.'.  In  any  triangle  the  square  on  a  side  opposite  an  acute 
angle  is  less  than  the  sum  of  the  sqicares  upon  the  other  two 
sides  by  twice  the  rectangle  ott  one  of  these  sides  and  the  pro- 
jection of  the  other  side  upon  it. 

3.  Let  the  angle  A  become  obtuse.  Then  D,  the  foot  of 
the  altitude  to  b,  passes  beyond  A, 
and  c^  changes  sign. 

a  .•.  [=1(5^1  changes  sign,         (161°) 

and  a^  =  b'^'-Vc^^ibc^. 

.'.  The  square  on  the  side  opposite 
obttcse  angle  i7i  an  obtuse- 
angled  triaftgle  is  greater  than  the  sum  of  the  squares  on  the 
other  two  sides  by  twice  the  rectangle  on  one  of  these  sides  and 
the  projection  of  the  other  side  upon  it. 


AREAL   RELATIONS.  125 

The  results  of  2  and  3  are  fundamental  in  the  theory  of 
triangles. 

These  results  are  but  one  ;  for,  assuming  as  we  have  done 
that  the  nn^Ci  is  to  be  subtracted  from  d'^  +  c^  when  A  is  an 
acute  angle,  the  change  in  sign  follows  necessarily  when  A 
becomes  obtuse,  since  in  that  case  the  en  changes  sign  because 
one  of  its  sides  changes  sign  (161°);  and  in  conformity  to 
algebraic  forms  - (  —  2bc-^=  +  2bc-^. 

Cor.  If  the  sides  a,  b,  c  of  a.  triangle  be  given  in  numbers, 

we  have  from  2  Ci~ , 

20 

which  gives  the  projection  of  c  on  k 

If  ^1  is  +  the  lA  is  acute  ; 
if  c^  is  o  the  z_A  is  ~~\ ; 
and  if  ^1  is  -  the  lA  is  obtuse. 

Ex.  The  sides  of  a  triangle  being  12,  13,  and  4,  to  find  the 
character  of  the  angle  opposite  side  13. 

Let  13  =  ^,  and  denote  the  other  sides  as  you  please,  ^.^., 
^=12  and  c=4.     Then 

.  _I2H4^-I3^_     3 

1 :r, ~  Q' 

24  6 

and  the  angle  opposite  side  13  is  obtuse. 

173°.   Theorem. — The  sum  of  the  squares  on  any  two  sides 
of  a  triangle  is  equal  to  twice  the  sum  of  the  squares  on  one- 
half  the  third  side  and  on  the  median  to  b 
that  side. 

BE  is  the  median  to  AC.     Then 
AB2  +  BC2=2(AE2  +  EB2). 

Proof.— l^^t  D  be  the  foot  of  the  altitude  ^  ^      ^ 

on  AC.     Consider  the  AABE  obtuse-angled  at  E,  and 

AB2  =  AE2-fEB2  +  2AE.ED.  (172°,  3) 

Next,  consider  the  ACBE  acute-angled  at  E,  and 

BC2  =  EC2-HEB2-2EC.ED.  (172°,  2) 


126  SYNTHETIC   GEOMETRY. 

Now,  adding  and  remembering  that  AE  =  EC, 

AB2  +  BC2  =  2AE2  +  2EB2.  q.e.d. 

Cor.  I.  Denoting  the  median  by  in  and  the  side  upon  which 
it  falls  by  b^  we  have  for  the  length  of  the  median 

4 
Cor.  2.  All  the  sides  of  an  equilateral  triangle  are  equal 
and  the  median  is  the  altitude  to  the  base  and  the  right 
bisector  of  the  base.  (53",  Cors.  2,  3) 

.'.  in  an  equilateral  triangle, 

ni^=p^  =  \aP-^  ox  p  =  \asj 2i^  a  being  the  side. 

173°.  Theorem. — The  sum  of  the  squares  on  the  sides  of  a 
quadrangle  is  equal  to  the  sum  of  the 
squares  on  the  diagonals,  and  four  times 
the  square  on  the  join  of  the  middle  points 
of  the  diagonals. 

E,  F  are  middle  points  of  AC  and  BD. 
Then         2(AB2)  =  AC2  +  BD2+4EF2. 

D  Proof.— ]6m  AF  and  CF. 

Then  AF  is  a  median  to  AABD,  and  CF  to  ACRD. 

AB2  +  AD2=2BF2-f2AF2,  (172^^ 

and  BC2  +  CD2  =  2BF2  +  2CF2, 

.-.  adding,  2(AB2)  =  4BF2  +  2(AF2+CF2). 

But  EF  is  a  median  to  AAFC. 

AF2  +  CF2  =  2CE2  +  2EF2,  (172^) 

2;(  AB2)  =  4BF2  +  4CE2  +  4EF2 

=  BD2  +  AC2  +  4EF2.  q.e.d. 

Since  squares  only  are  involved  this  relation  is  true  with- 
out any  modification  for  all  quadrangles. 

Cor.  I.  When  the  quadrangle  becomes  a  1  7  the  diagonals 
bisect  one  another  (81°,  3)  and  EF  becomes  zero. 

.•.  the  sum  of  the  squares  on  the  sides  of  a  parallelogram 
is  equal  to  the  sum  of  the  squares  on  its  diagonals. 


AREAL   RELATIONS.  I  2/ 

174°.  Let  ABC  be  an  isosceles  triangle  and  P  be  any  point 
in  the  base  AC,  and  let  D  be  the  middle  point  of  ^ 

the  base,  and  therefore  the  foot  of  the  altitude. 
In  the  ABAP,  acute-angled  at  A, 

BP2=BA2  +  AP2-2AP.AD,      (172°,  2)      / 
BA2-BP2=AP(2AD-AP)  / 

=  AP.PC.  Q     A    P     D        C 

If  P  moves  to  Q,  AP  becomes  AQ  and  changes  sign,  BP 
becomes  BQ  which  is  >  BA,  and  thus  both  sides  of  the 
equality  change  sign  together  as  they  pass  through  zero  by 
P  passing  A. 

Now,  of  the  two  segments  from  B  we  always  know  which 
is  the  greater  by  63°,  and  if  we  write  PA  for  AP  the  izdPA  .  PC 
is  positive  when  P  is  on  the  Q  side  of  A.  Hence,  considering 
the  rectangle  as  being  always  positive,  we  may  state  the 
theorem — 

The  differoice  between  the  squares  on  a  side  of  an  isosceles 
triangle  and  on  the  join  of  the  vertex  to  any  point  in  the  base 
is  equal  to  the  rectangle  on  the  segments  into  which  that poijit 
divides  the  base. 

175°.  I.  From  174°  we  have  BA2-BP2  =  AP .  PC.  Now 
BA  is  fixed,  therefore  the  [=]AP .  PC  increases  as  BP  de- 
creases. But  BP  is  least  when  P  is  at  D  (63°,  i),  therefore 
the  cdAP  .  PC  is  greatest  when  P  is  at  D. 

Def  I. — A  variable  magnitude,  which  by  continuous 
change  may  increase  until  a  greatest  value  is  reached  and 
then  decrease,  is  said  to  be  capable  of  a  maximum,  and  the 
greatest  value  reached  is  its  maximum. 

Thus  as  P  moves  from  A  to  C  the  imAP .  PC  increases 
from  zero,  when  P  is  at  A,  to  its  maximum  value,  when  P  is 
at  D,  and  then  decreases  again  to  zero,  when  P  comes  to  C. 

And  as  AC  may  be  considered  to  be  any  segment  divided 
at  P, 

.'.  The  maximum  rectangle  on  the  parts  of  a  given  segment 
is  formed  by  bisecting  the  segment; 


128  SYNTHETIC   GEOMETRY. 

Or,  of  all  rectangles  with  a  given  perimeter  the  square  has 
the  greatest  area. 

2.         AC2=(AP  +  PC)2  =  AP2  +  PC2  +  2AP.PC       (166°,  i) 
=  AP2  +  PC2  +  2(AB2-BP2).  (174°) 

But  AC  and  AB  are  constant, 

AP2  +  PC2  decreases  as  BP2  decreases. 
But  BP  is  least  when  P  is  at  D, 

AP2  +  PC2  is  least  when  P  is  at  D. 
Def.  2. — A  variable  magnitude  which  by  continuous  change 
decreases  until  it  reaches  a  least  value  and  then  increases  is 
said  to  be  capable  of  a  miniinnm,  and  the  least  value  attained 
is  called  its  miniimiin. 

.'.  The  sum  of  the  squares  on  the  ttvo  parts  of  a  given  seg- 
ment is  a  minimum  when  the  segme7it  is  bisected. 

I75j°.  The  following  examples  give  theorems  of  importance. 
B  Ex.  I.  Let  ABC  be  any  triangle  and  BD  the 

altitude  to  side  b.     Then 


y    p\  \a  .      b-'  +  C'^-a- 


^1= zi '  (172°,  Cor.) 

A       fa  D    c  But      P^^c^-c^^-ic-^-c^ic-c^, 

and  /\  =  \bp.  (153°,  2) 

Now,  .+.,  =  (i±f)^^=(A±i±^(^±fZ-).^ 

2b  lb 

and  c-c  _a''-{b-cf_{a  +  b-c){a-b  +  c) 

'^  2b  2b  ' 


)'^=:i6/\^  =  {a  +  b  +  c){b^-c-a){c+a-b){a  +  b-c\ 
and   by  writing  s  for  \{a  +  b  +  c),  dindi  accordingly  s-a   for 
\{b  +  c—a),  etc.,  we  obtain 

A = slXs-a){s-b){s~c): 
This  important  relation  gives  the  area  of  the  A  in  terms  of 
its  three  sides. 

Ex.  2.  Let  ABC  be  an  equilateral  A     Then  the  area  may 

be  found  from  Ex.  i  by  making  a  =  b  =  c,  when  the  reduced 

2 
expression  becomes,  A  =  '^  \^3- 

4 


AREAL    RELATIONS.  1 29 

Ex.  3.  To  find  the  area  of  a  regular  octagon  in  terms  of  its 
circumradius.  A  d 

Let  A,  B,  C  be  three  vertices  of  the  octagon 
and  O  the  centre.  Complete  the  square  OD, 
and  draw  BE  ±  to  OA. 

Since  lEOB^^~],  and  OB  =  r, 
E0  =  EB-|rV2, 
and       A0AB=^0A.EB=|r.|rV2=irV2. 
But  AOAB  is  one-eighth  of  the  octagon, 
Oct.  =  2rV2. 

Exercises. 

1.  ABC  is  right-angled  at  B,  and  E  and  F  are  middle  points 

of  BA  and  BC  respectively.   Then  5AC2  =  4(CE2-^  AF2). 

2.  ABC  is  right-angled  at  B  and  O  is  the  middle  of  AC, 

and   D   is   the  foot   of  the   altitude  from  B.     Then 
2AC.OD  =  AB2-BC2. 

3.  ABC  is  right-angled  at  B  and,  on  AC,  AD  is  taken  equal 

to  AB,  and  on  CA,  CE  is  taken  equal  to  CB.     Then 
ED2=2AE.DC. 

4.  The  square  on  the  sum  of  the  sides  of  a  right-angled  tri- 

angle exceeds  the  square  on  the  hypothenuse  by  twice 
the  area  of  the  triangle. 

5.  To  find  the  side  of  a  square  which  is  equal  to  the  sum  of 

two  given  squares. 

6.  To  find  the  side  of  a  square  which  is  equal  to  the  differ- 

ence of  two  given  squares. 

7.  The  equilateral  triangle  described  upon  the  hypothenuse 

of  a  right-angled  triangle  is  equal  to  the  sum  of  the 
equilateral  triangles  described  on  the  sides. 

8.  ABC  is  a  triangle  having  AB  =  CB,  and  AD  is  J_  upon 

BC.     Then  AC2=2CB.CD. 

9.  Four  times  the  sum  of  the  squares  on  the  three  medians 

of  a  triangle  is  equal  to  three  times  the  sum  of  the 
squares  on  the  sides. 

I 


I30  SYNTHETIC   GEOMETRY. 

10.  ABCD  is  a  rectangle  and  P  is  any  point.     Then 

PA2  +  PC2=PB2  +  PD2. 

11.  O  is  the  centre  of  a  circle,  and  AOB  is  a  centre-line. 

OA=-OB  and  C  is  any  point  on  the  circle.     Then 
AC2  +  BC2  =  a  constant. 

Define  a  circle  as  the  locus  of  the  point  C. 

12.  AD  is  a  perpendicular  upon  the  line  OB,  and  BE  is  a 

perpendicular    upon   the  line   OA.      Then    OA .  OE 
=  OB.OD. 

13.  Two  equal  circles  pass  each  through  the  centre  of  the 

other.     If  A,  B  be  the  centres  and  E,  F  be  the  points 
of  intersection,  EF2  =  3AB2. 

If  EA  produced  meets  one  circle  in  P  and  AB  pro- 
duced meets  the  other  in  Q,  PQ2  =  7AB2. 

14.  ABC  is  a  triangle  having  the  angle  A  two-thirds  of  a 

right  angle.     Then  AB2-fAC2  =  BC2  +  AC.  AB. 

15.  In  the  triangle  ABC,  D  is  the  foot  of  the  altitude  to  AC 

and  E  is  the  middle  point  of  the  same  side.     Then 
2ED.AC  =  AB2-BC2. 

16.  AD  is  a  line  to  the  base  of  the  triangle  ABC,  and  O  is 

the  middle  point  of  AD.     If  AB2  +  BD2=AC2  +  CD2, 
then  OB  =  OC. 

17.  ABC  is  right-angled  at  B  and  BD  is  the  altitude  to  AC. 

Then  AB.  CD  =  BD  .  BC  and  AD  .  CB  =  BA.  BD. 

18.  ABC  is  a  triangle  and  OX,  OY,  OZ  perpendiculars  from 

any  point  O  on  BC,  CA,  and  AB  respectively.     Then 
BX2  -f-  C  Y2  -f  AZ2  =  CX2  +  A  Y2  -1-  BZl 
A  similar  relation  holds  for  any  polygon. 

19.  AA^,  BBi  are  the  diagonals  of  a  rectangle  and  P  any  point. 

Then  PA2+ PB2  +  PAi2  +  PBi2=AAj--}-4P02,  where  O 
is  the  intersection  of  the  diagonals. 

20.  ABC  is  a  triangle,  AD,  BE,  CF  its  medians,  and  P  any 

point.     Then 
PA2-hPB2  +  PC2  =  PD2-fPE2+PF2-fi(AD-  +  BE2-fCF2), 
or  2PA2  =  2PD2-i- 12;a22^  where  in  is  a  median. 

21.  If  O  be  the  centroid  in  20, 


AREAL   RELATIONS.  I31 

PA2+PB2  +  PC2  =  3P02  +  MAD2  +  BE2  +  CF2), 
or  SPA2  =  3P02  +  fS;/z2. 

22.  ABCD  is  a  square  and  AA',  BE',  CC,  DD'  perpendicu- 

lars upon  any  line  L.     Then 

(AA'2  +  CC'2)~2BB'.  DD'  =  area  of  the  square. 

23.  The  sum  of  the  squares  on  the  diagonals  of  any  quad- 

rangle is  equal  to  twice  the  sum  of  the  squares  on  the 
joins  of  the  middle  points  of  opposite  sides. 

24.  ABCD  is  a  trapezoid  having  AD  parallel  to  BC.     Then 

AB2  +  CD2  +  2AD.BC  =  AC2  +  BD2. 

25.  If  A,  B,  C  be  equidistant  points  in  line,  and  D  a  fourth 

point  in  same  line,  the  difference  between  the  squares 
on  AB  and  DB  is  equal  to  the  rectangle  on  AD  and 
CD. 

26.  If  A,  B,  C,  D  be  any  four  points  in  line, 

AD2  +  BC2=:AC2-fBD2  +  2AB.CD. 

27.  Any  rectangle  is  equal  to  one-half  the  rectangle  on  the 

diagonals  of  the  squares  described  on  adjacent  sides. 

28.  In  the  triangle  ABC,  D  is  any  point  in  BC,  E  is  the 

middle  point  of  AC  and  F  of  BC.     Then 
AB24-AC2  =  AD2  +  4EF2-f-2BD.DC. 

29.  The  sides  of  a  rectangle  are  a  and  b.     If/  be  the  length 

of  the  perpendicular  from  a  vertex  upon  a  diagonal  and 
q  be  the  distance  between  the  feet  of  the  two  parallel 
perpendiculars  so  drawn, 

ps!'cfiV^=  ab  and  q^'c^T¥=  b^ -a^  {b  >  a), 
what  line-segment  is  denoted  by  s/ii^  +  b'^  ? 

30.  ABCD  is  a  square.     P  is  a  point  in  AB  produced,  and  Q 

is  a  point  in  AD.  If  the  rectangle  BP .  QD  is  con- 
stant, the  triangle  PQC  is  constant. 

31.  If  the  lengths  of  the  sides  of  a  triangle  be  expressed  by 

x^+i,  x^-  I,  and  2r,  the  triangle  is  right-angled. 

32.  If  a  and  c  be  the  sides  of  a  right-angled  triangle  and  p  be 

the  altitude  to  the  hypothenuse, 

14-1=1. 

^2   •   ^2      ^2 


132  SYNTHETIC   GEOMETRY. 

33.  The   triangle  whose  sides   are  20,   15,  and  12  has  an 

obtuse  angle. 

34.  The  area  of  an  isosceles  triangle  is  8^15  and  the  side  is 

twice  as  long  as  the  base.  Find  the  length  of  the  side 
of  the  triangle. 

35.  What  is  the  length  of  the  side  of  an  equilateral  triangle 

which  is  equal  to  the  triangle  whose  sides  are  13,  14, 
and  15? 

36.  If  AB  is  divided  in  C  so  that  AC2-2BC2,  then 

AB2  +  BC2=2AB.AC. 

37.  Applying  the  principle  of  continuity  state  the  resulting 

theorem  when  B  comes  to  D  in  (i)  the  Fig.  of  172°, 
(2)  the  Fig.  of  173°. 

38.  Applying  the  principle  of  continuity  state  the  resulting 

theorem  when  B  comes  to  E  in  the  Fig.  of  173°. 

39.  The  bisector  of  the  right  angle  of  a  right-angled  triangle 

cuts  the  hypothenuse  at  a  distance  a  from  the  middle 
point,  and  the  hypothenuse  is  lb.  Find  the  lengths  of 
the  sides  of  the  triangle. 

40.  Construct  an  equilateral  triangle  having  one  vertex  at  a 

given  point  and  the  remaining  vertices  upon  two  given 
parallel  lines. 

41.  A  square  of  cardboard  whose  side  is  s  stands  upright 

with  one  edge  resting  upon  a  table.  If  a  lower  corner 
be  raised  vertically  through  a  distance  a^  through 
what  distance  will  the  corner  directly  above  it  be 
raised  ? 

42.  What  would  be  the  expression  for  the  area  of  a  rectangle 

if  the  area  of  the  equilateral  triangle  having  its  side 
the  u.l.  were  taken  as  the  u.a?. 

43.  The  opposite  walls  of  a  house  are  12  and  16  feet  high  and 

20  feet  apart.     The  roof  is  right-angled  at  the  ridge 
and  has  the  same  inclination  on  each  side.     Find  the 
lengths  of  the  rafters. 
44    Two  circles  intersect  in  P  and  Q.     The  longest  chord 
through  P  is  perpendicular  to  PQ. 


AREAL  RELATIONS. 


33 


45.  The  largest  triangle  with  a  given  perimeter  is  an  equi- 

lateral triangle. 

46.  The  largest  triangle  having  its  base  and  the  sum  of  the 

other  two  sides  given  is  isosceles. 

47.  The  largest  polygon  of  given  species  and  given  perimeter 

is  regular. 

48.  The  largest  isosceles  triangle  with  variable  base  has  its 

sides  perpendicular  to  one  another, 

49.  The  largest  rectangle  inscribed  in  an  acute-angled  tri- 

angle and  having  one  side  lying  on  a  side  of  the 
triangle  has  its  altitude  one-half  that  of  the  triangle. 

50.  L,  M  are  two  lines  meeting  in  O,  and  P  is  any  point. 

APB  is  a  variable  hne  cutting  L  in  A  and  M  in  B. 
The  triangle  AOB  is  least  when  P  bisects  AB. 


EQUALITIES  OF  RECTANGLES  ON  SEGMENTS 
RELATED  TO  THE  CIRCLE. 


176°.  Theorem. — If  two  secants  to  the  same  circle  inter- 
sect, the  rectangle  on  the  segments  between  the  point  of 
intersection  and  the  circle  with  respect  to  one  of  the  secants 
is  equal  to  the  corresponding  rectangle  with  respect  to  the 
other  secant. 

I.  Let  the  point  of  intersection  be 
within  the  circle.     Then 

AP.PB  =  CP.PD. 

Proof. — AOB  is  an  isosceles  triangle, 
and  P  is  a  point  on  the  base  AB. 

OA2-OP2  =  AP.PB.       (174°) 

Similarly,  COD  is  an  isosceles  tri- 
angle, and  P  a  point  in  the  base  CD, 

OC2-OP2=CP.PD. 
But  OC  =  OA, 

AP.PB  =  CP.PD.  q.e.d. 


134 


SYNTHETIC   GEOMETRY. 


Cor.  I.  (a)  Let  CD  become  a  diameter  and  be  _L  to  AB. 

Then     AP.  PB  becomes  AP2,    (96°,  Cor.  5) 

AP2  =  CP.PD, 
and  denoting  AP  by  c,  CP  by  -z/,  and  the 
radius  of  the  circle  by  r,  this  becomes 

which  is  a  relation  between  a  chord  of  a  0, 
the  radius  of  the  0,  and  the  distance  CP, 
commonly  called  the  versed  sine,  of  the  arc  AB. 

{b)  When  the  point  of  intersection  P  passes  without  the  0 
we  have  still,  by  the  principle  of  con- 
tinuity, AP .  PB  =  CP .  PD.  But  the  as 
being  now  both  negative  we  make  them 
both  positive  by  writing 

PA.PB  =  PC.PD. 


Cor.  2.  When  the  secant  PAB  be- 
comes the  tangent  PT  (109°),  A  and 
B  coincide  at  T,  and  PA.PB  becomes 
PT2,  .-.  PT2  =  PC.PD, 

.e.,  if  a  tangent  and  a  secant  be  drawn 
from  the  same  point  to  a  circle^  the  square  on  the  tangent  is 
equal  to  the  rectangle  on  the  segments  of  the  secant  between 
the  point  a7id  the  circle. 

Cor.  3.  Conversely,  if  T  is  on  the  circle  and  PT2=PC .  PD, 
PT  is  a  tangent  and  T  is  the  point  of  contact. 

For,  if  the  line  PT  is  not  a  tangent  it  must  cut  the  circle  in 
some  second  point  T'  (94°).     Then 

PT.PT'  =  PC.PD  =  PT2. 
Therefore  PT  =  PT',  which  is  not  true  unless  T  and  T'  coin- 
cide.    Hence  PT  is  a  tangent  and  T  is  the  point  of  contact. 

Cor.  4.  Let  one  of  the  secants  become  a  centre-line  as 
PEF.  Denote  PT  by  /,  PE  by  h,  and  the  radius  of  the  circle 
byr.     Then  PT2  =  PE.PF 

becomes  f^—hip-r+h). 


AREAL   RELATIONS.  1 35 


Exercises. 


1.  The  shortest  segment  from  a  point  to  a  circle  is  a  portion 

of  the  centre-line  through  the  point. 

2.  The  longest  segment  from  a  point  to  a  circle  is  a  portion 

of  the  centre-line  through  the  point. 

3.  If  two  chords  of  a  circle  are  perpendicular  to  one  another 

the  sum  of  the  squares  on  the  segments  between  the 
point  of  intersection  and  the  circle  is  equal  to  the 
square  on  the  diameter. 

4.  The  span  of  a  circular  arch  is  120  feet  and  it  rises  15 

feet  in  the  middle.  With  what  radius  is  it  con- 
structed? 

5.  A  conical  glass  is  b  inches  deep  and  a  inches  across  the 

mouth.  A  sphere  of  radius  r  is  dropped  into  it.  How 
far  is  the  centre  of  the  sphere  from  the  bottom  of  the 
glass  ? 

6.  The  earth's  diameter  being  assumed  at  7,960  miles,  how 

far  over  its  surface  can  a  person  see  from  the  top  of  a 
mountain  3  miles  high  ? 

7.  How  much  does  the  surface  of  still  water  fall  away  from 

the  level  in  one  mile  ? 

8.  Two  circles  whose  radii  are  10  and  6  have  their  centres 

12  feet  apart.  Find  the  length  of  their  common  chord, 
and  also  that  of  their  common  tangent. 

9.  Two  parallel  chords  of  a  circle  are  c  and  c-^  and  their 

distance    apart    is    d^    to    find    the    radius    of    the 
circle. 
10.  If  V  is  the  versed  sine  of  an  arc,  k  the  chord  of  half  the 
arc,  and  r  the  radius,  k^  =  2vr. 

177°.  Theorem. — If  upon  each  of  two  intersecting  lines  a 
pair  of  points  be  taken  such  that  the  rectangle  en  the  seg- 
ments between  the  points  of  intersection  and  the  assumed 
points  in  one  of  the  lines  is  equal  to  the  corresponding  rect- 


1.^6 


SYNTHETIC  GEOMETRY. 


angle  for  the  other  Hne,  the  four  assumed  points  are  concyclic. 

(Converse  of  176°.) 
L  and  M  intersect  in  O,  and 
OA.OB  =  OC.OD. 
Then  A,  B,  C,  and  D  are  concydic. 

Proof. — Since  the  ens  are  equal,  if  A  and  B  lie  upon  the 
same  side  of  O,  C  and  D  must  lie  upon  the  same  side  of  O  ; 
and  if  A  and  B  He  upon  opposite  sides  of  O,  C  and  D  must 
lie  upon  opposite  sides  of  O. 

Let  a  0  pass  through  A,  B,  C,  and  let  it  cut  M  in  a  second 
point  E.     Then  OA.OB  =  OC.OE.  (176°) 

But  OA.OB  =  OC.OD.  (hyp.) 

OD  =  OE, 
and  as  D  and  E  are  upon  the  same  side  of  O  they  must  co- 
incide; .*.     A,  B,  C,  D  are  concyclic.  q.e.d. 


178°.  Let  two  circles  excluding  each  other  without  contact 
have  their  centres  at  A  and  B,  and  let  C  be  the  point,  on 
their  common  centre-line>  which  divides  AB  so  that  the 
difference  between  the  squares  on  the  segments  AC  and  CB 
is  equal  to  the  difference  between  the  squares  on  the  con- 
terminous radii.  Through  C 
draw  the  line  PCD  X  to  AB, 
and  from  any  point  P  on  this 
line  draw  tangents  PT  and  PT' 
to  the  circles. 

Join  AT  and  BT'. 
Then,  by  construction, 

AC2-BC2  =  AT2-Br2. 

But,  since  PC  is  an  altitude  in 

BC2=AP2-BP2,  (172°,  i) 

AP2=AT2  +  PT2, 

BP2  =  BT'2  +  PT'2,        (169°,  Cor.  i) 

p'J'2_p'p'2 

PT  =  PT'. 


AREAL   RELATIONS. 


137 


Therefore  PCD  is  the  locus  of  a  point  from  which  equal 
tangents  are  drawn  to  the  two  circles. 

Def. — This  locus  is  called  the  radical  axis  of  the  circles, 
and  is  a  line  of  great  importance  in  studying  the  relations  of 
two  or  more  circles. 

Cor.  I.  The  radical  axis  of  two  circles  bisects  their  com- 
mon tangents. 

Cor.  2.  When  two  circles  intersect,  their  radical  axis  is 
their  common  chord. 

Cor.  3.  When  two  circles  touch  externally,  the  common 
tangent  at  the  point  of  contact  bisects  the  other  common 
tangents. 

179°.  The  following  examples  give  theorems  of  some  im- 
portance. 

Ex.  I.  P  is  any  point  without  a  circle  and  TT'  is  the  chord 

of  contact  (114°,  Def.)  for  the  point  P. 

TT'   cuts   the   centre-Hne   PO   in    O. 

Then,  PTO  being  a  "]»  (i  10°) 

OQ.OP  =  OT2.  (169°) 

.*.  the  radius  is  a  geometric  mean  be- 
tween the  join  of  any  poifit  with  the 
centre  and  the  perpendictdar  from  the 
centre  upon  the  chord  of  contact  of  the  point. 

Def. — P  and  Q  are  called  inverse  points  with  respect  to 
the  circle. 

Ex.  2.  Let  PQ  be  a  common  direct  tangent  to  the  circles 
having  O  and  O'  as  centres. 

Let  OP  and  O'Q  be  radii 
to  the  points  of  contact,  and 
let  QR  be  II  to  00'.  Denote 
the  radii  by  r  and  /.  Then 
AC  =  00'  +  r-r', 
BD  =  00'-r-l-r'. 
.-.  AC.BD  =  00'2-(r-r')2  = 


QR2-PR2  =  PQ2.  (169°,  Cor.  i) 


138  SYNTHETIC  GEOMETRY. 

Similarly  it  may  be  shown  that 
AD  .  BC  =  square  on  the  transverse  common  tangent. 


Exercises. 

1.  The  greater  of  two  chords  in  a  circle  is  nearer  the  centre 

than  the  other. 

2.  Of  two  chords  unequally  distant  from  the  centre  the  one 

nearer  the  centre  is  the  greater. 

3.  AB  is  the  diameter  of  a  circle,  and  P,  Q  any  two  points  on 

the  curve.  AP  and  BQ  intersect  in  C,  and  AQ  and  BP 
in  C.     Then 

AP.AC  +  BQ.BC  =  AC'.AQ  +  BC'.BP. 

4.  Two  chords  of  a  circle,  AB  and  CD,  intersect  in  O  and 

are  perpendicular  to  one  another.     If  R  denotes  the 
radius  of  the  circle  and  E  its  centre, 
8R2  =  AB2  +  CD2  +  40E2. 

5.  Circles  are  described  on  the  four  sides  of  a  quadrangle  as 

diameters.  The  common  chord  of  any  two  adjacent 
circles  is  parallel  to  the  common  chord  of  the  other 
two. 

6.  A  circle  S  and  a  line  L,  without  one  another,  are  touched 

by  a  variable  circle  Z.  The  chord  of  contact  of  Z  passes 
through  that  point  of  S  which  is  farthest  distant 
from  L. 

7.  ABC  is  an  equilateral  triangle  and  P  is  any  point  on  its 

circumcircle.  Then  PA  +  PB  +  PC=o,  if  we  consider 
the  line  crossing  the  triangle  as  being  negative. 

8.  CD  is  a  chord  parallel  to  the  diameter  AB,  and  P  is  any 

point  in  that  diameter.     Then 

PC2  +  PD2  =  PA2  +  PB2. 


CONSTRUCTIVE  GEOMETRY.  1 39 

SECTION   V. 

CONSTRUCTIVE    GEOMETRY. 

1 80°.  Problem.— PC&  being  a  given  segment,  to  construct 
the  segment  ABv/2.  A b 

C^;zj/r.— Draw  BC  JL  to  AB  and  equal  to  it. 
Then  AC  is  the  segment  ABx/2. 

Proof. — Since  ABC  is  right-angled  at  B, 

AC2=AB2  +  BC2  =  2AB2,   (169,  Cor.  i) 
AC  =  ABV2. 
Cor.  The  square  on  the  diagonal  of  a  given  square  is  equal 
to  twice  the  given  square. 

181°.  Problem.— To  construct  ABv'S- 
Constr. — Take  BC  in  line  with  AB  and  equal 
to  it,  and  on  AC  construct  an  equilateral  tri- 
angle ADC.  (124°,  Cor.  I) 
BD  is  the  segment  ABV3. 

Proof.— KBJ)  is  a  ~1,  and  AD  =  AC  =  2AB. 
Also  AD2  =  AB2-hBD2  =  4AB2.         (169°,  Cor.  i) 

BD2  =  3AB2,  and  BD^AB^/S- 

Cor.  Since  BD  is  the  altitude  of  an  equilateral  triangle 
and  AB  is  one-half  the  side, 

.'.  the  square  on  the  altitude  of  an  equilateral  triangle  is 
equal  to  three  times  the  square  on  the  half  side. 

182°.  Problem.— To  construct  AB^S- 
Constr.  —Draw  BC  X  to  AB  and  equal  to  twice 
AB.     Then  AC  is  the  segment  ABV5. 
Pr^^— Since  Z.B  is  a  right  angle, 
AC2  =  ABHBC2. 
But  BC2  =  4AB2; 

AC2=5AB2, 
and  AC  =  ABV5. 


140 


SYNTHETIC    GEOMETRY. 


183°.  The  three  foregoing  problems  furnish  elements  of 
construction  which  are  often  convenient.  A  few  examples 
are  given. 

Ex.  I.  AB  being  a  given  segment,  to  find  a  point  C  in  its 
,  line  such  that  AC'  =  AB  .  CB. 

A  c  B     Analysis-  AC2  =  AB.  CB  =  AB(AB- AC), 

AC^  +  AC.AB  =  AB2. 
Considering  this  an  algebraic  form  and  solving  as  a  quad- 
ratic in  AC,  we  have        AC  =  J(ABV5  -  AB), 
and  this  is  to  be  constructed. 

Constr. — Construct  AD=ABV5  (by  182°)  as  in  the  figure, 

c ' A  9      ^ ^^^    ^^t    E    be    the 

middle  point  of  BD. 

TakeDF  =  DE. 
Then 

AF=ABV5-AB; 
.*.  bisecting  AF  in  G, 
AG  =  AC 

=MABV5-AB), 
and  the  point  C  is 
found. 

Again,  since  ^J^  has 
two    signs    +    or   -, 
Jy^'  take  its  negative  sign 
and  we  have  AC  =  -  ^AB^S  +  AB). 

Therefore,    for    the    point    C,    on    AD    produced    take 
DF'  =  DE,  and  bisect  AF'  in  G'.     Then 
AG'=|(ABV5  +  AB); 
and  since  AC  is  negative  we  set  off  AG'  from  A  to  C,  and 
C  is  a  second  point. 

The  points  C  and  C  satisfy  the  conditions, 

AC2  =  AB  .  CB  and  AC2=AB  .  CB. 
A  construction  effected  in  this  way  requires  no  proof  other 
than  the  equation  which  it  represents. 


\ 


CONSTRUCTIVE   GEOMETRY.  I4I 

It  is  readily  proved  however.     For 
AD2=5AB-,  and  also  AD2=(AF  +  FD)2  =  (2AC  +  AB)2, 
whence  AC2  =  AB(AB  -  AC) = AB .  CB. 

It  will  be  noticed  that  the  constructions  for  finding  the  two 
points  differ  only  by  some  of  the  segments  being  taken  in 
different  senses.  Thus,  for  C,  DE  is  taken  from  DA,  and  for 
C\  added  to  DA ;  and  for  C,  AC  is  taken  in  a  positive  sense 
equal  to  AG,  and  for  C,  AC  is  taken  in  a  negative  sense 
equal  to  AG'. 

In  connection  with  the  present  example  we  remark  : — 

1.  Where  the  analysis  of  a  problem  involves  the  solution  of 
a  quadratic  equation,  the  problem  has  two  solutions  corre- 
sponding to  the  roots  of  the  equation. 

2.  Both  of  the  solutions  may  be  applicable  to  the  wording 
of  the  problem  or  only  one  may  be. 

3.  The  cause  of  the  inapplicability  of  one  of  the  solutions 
is  commonly  due  to  the  fact  that  a  mathematical  symbol  is 
more  general  in  its  significance  than  the  words  of  a  spoken 
language. 

4.  Both  solutions  may  usually  be  made  applicable  by  some 
change  in  the  wording  of  the  problem  so  as  to  generalize  it. 

The  preceding  problem  may  be  stated  as  follows,  but 
whether  both  solutions  apply  to  it,  or  only  one,  will  depend 
upon  our  definition  of  the  word  "  part."     See  Art.  23°. 

To  divide  a  given  segment  so  that  the  square  tipon  one  of 
the  parts  is  equal  to  the  rectangle  on  the  whole  segment  and 
the  other  part. 

Def. — A  segment  thus  divided  is  said  to  be  divided  into 
extreme  and  mea7i  ratio,  or  in  median  section. 

Ex.  2.  To  describe  a  square  when  the  sum  of  its  side  and 
diagonal  is  given. 

Analysis. — If  AB  is  the  side  of  a  square,  AB^/2  is  its 
diagonal,  (iSo°) 


142 


SYNTHETIC   GEOMETRY. 


yH' 


.'.  AB(i  +J2)  is  a  given  segment  =S,  say.     Then 

AB  =  S(V2-i). 

Constr. — Let  EF  be  the  given  segment  S. 
Draw  FG  JL  and   =   to  EF,  and   with 
centre  G  and  radius  GF  describe  a  0  cut- 
ting EG  in  H  and  H'. 

EH  is  the  side  of  the  square;  whence  the 
square  is  easily  constructed. 

If  we  enquire  what  EH'  means,  we  find  it  to  be  the  side  of 
the  square  in  which  the  difference  between  the  side  and 
diagonal  is  the  given  segment  S.  The  double  solution  here 
is  very  suggestive,  but  we  leave  its  discussion  to  the  reader. 

184°.  Problem. — To  find  a  segment  such  that  the  rectangle 
on  it  and  a  given  segment  shall  be  equal  to  a  given  rectangle. 

p  G  E 

"■ Constr. — Let  S  be  the  given  segment, 

and  AC  the  given  rectangle. 

On  DA  produced  make  AP  =  S,  and 
draw  PBO  to  cut  DC  produced  in  Q. 
CQ  is  the  segment  required. 

-Complete  the  as  PEQD,  PGBA,  and  BCQF. 
i=iAC=[=:GF  =  GB.BF  =  PA.CQ, 
S.CQ  =  [=iAC. 

Def. — The  segments  AP  and  CQ  are  reciprocals  of  one 
another  with  respect  to  the  cuAC  as  unit. 


185°.  Problem. — To   find  the   side   of  a  square  which  is 
equal  to  a  given  rectangle. 

Constr. — Let    AC    be    the    rectangle. 
Make   BE  =  BC   and   in   line  with  BA. 
On  AE  describe  a  semicircle,  and  pro- 
duce CB  to  meet  it  in  F. 
^  ^  BF  is  the  side  of  the  required  square. 

Proof. — Since  AE  is  a  diameter  and  FB  a  half  chord  JL  to  it, 


/"^ 

r\ 

^ 

B 

CONSTRUCTIVE   GEOMETRY. 


143 


BF2=AB.BE, 
BF2=AB.BC. 


(176°,  Cor.  i) 


Cor.  This  is  identical  with  the  problem,  "To  find  a  geo- 
metric mean  between  two  given  segments,"  and  it  furnishes 
the  means  of  constructing  the  segment  a,  when  a=^>Jbc,  b  and 
^  being  given.  (165°) 

Ex.  I.  To  construct  an  equilateral  triangle  equal  to  a  given 
rectangle.  b  c 

Let  AC  be  the  given  rectangle,  and  suppose 
PQR  to  be  the  required  triangle.     Then 
AB.BC  =  iPR.QT 
=  PT.QT. 
But  QT  =  PTV3,  (181°,  Cor.) 

PT.QT=PTV3 
whence  PT2=ABV3-iBC.  p     t     r 

And  PT  is  the  side  of  a  square  equal  to  the  rectangle  whose 
sides  are  ABAy3  and  ^BC,  and  is  found  by  means  of  181°, 
127°,  and  185°. 
Thence  the  triangle  is  readily  constructed. 

Ex.  2.  To  bisect  the  area  of  a  triangle  by  a  line  parallel  to 

its  base. 

Let  ABC  be  the  triangle,  and  assume 

PQ  as  the  required  line,  and  complete 

the  parallelograms  AEBC,  KFBC,  and 

let  BD  be  the  altitude  to  AC.     Because 

PQ  is  II  to  AC,  BD  is  ±  to  PQ.     Now 

/=7EP  =  Z=Z7PC,  (145°) 

E=7FC  =  e:^EQ,  or  PQ.BD  =  AC.BG.       (153°,  i) 

But      2Z=7FQ  =  £Z:7EC,  or  2PQ.bg  =  AC.  BD; 

.*.  dividing  one  equation  by  the  other,  and  reducing  to  one 

line,  BD''i  =  2BG2; 

and  therefore  BG  is  one-half  the   diagonal    of  the   square 

of  which  BD  is  the  side,  and  the  position  of  PQ  is  de- 
termined. 


144 


SYNTHETIC   GEOMETRY. 


1 86°.  Problem. — To  find  the  circle  which  shall  pass  through 
two  given  points  and  touch  a  given  line. 
Let  A,  B  be  the  given  points  and  L 
the  given  line. 

Constr. — Let  the  line  AB  cut  L  in  O. 
Take  OP  =  OP',  a  geometric  mean  be- 
tween OA  and  OB  (185°).  The  circles 
through  the  two  sets  of  three  points  A,  B,  P  and  A,  B,  P'  are 
the  two  solutions. 

The  proof  is  left  to  the  reader.     (See  176°,  Cor.  2.) 

187°.  Problem. — To  find  a  0  to  pass  through  two  given 
points  and  touch  a  given  0. 

Let  A,  B  be  the  points  and  S 
the  given  0. 

Constr, — Through  A  and  B 
draw  any  0  so  as  to  cut  S  in 
two  points  C  and  D.  Let  the 
line  CD  meet  the  hne  AB  in  O. 
From  O  draw  tangents  OP  and 
OQ  to  the  0S  (114°).  Pand 
Q  are  the  points  of  contact  for 
the  0s  which  pass  through  A  and  B  and  touch  S.  There- 
fore the  0s  through  the  two  sets  of  three  points  A,  B,  P  and 
A,  B,  Q  are  the  0s  required. 

Proof.-        OB.OA=OC.OD  =  OQ2=OP2; 
therefore  the  0s  through  A,  B,  P  and  A,  B,  Q  have  OP  and 
OQ  as  tangents  (176°,  Cor.  3).     But  these  are  also  tangents 
to  0S  ;  therefore  P  and  Q  are  the  points  of  contact  of  the 
required  0s. 

Exercises. 

1.  Describe  a  square  that  shall  have  twice  the  area  of  a 

given  square. 

2.  Describe  an  equilateral  triangle  equal  to  a  given  square. 


CONSTRUCTIVE   GEOMETRY.  145 

3.  Describe  an  equilateral  triangle  having  five  times  the  area 

of  a  given  equilateral  triangle. 

4.  Construct  ABV7,  where  AB  is  a  given  segment. 

5.  Construct  sJa^  +  b^  and  sla^-b'^,  where  a  and  b  denote 

given  line  segments. 

6.  Divide  the  segment  AB  in  C  so  that  AC2  =  2CB2.     Show 

that  AC  is  the  diagonal  of  the  square  on  CB.  Does 
this  hold  for  external  division  also  ? 

7.  ABCD  is  a  rectangle  and  DE,  a  part  of  DA,  is  equal 

to  DC.  EF,  perpendicular  to  AD,  meets  the  circle 
having  A  as  centre  and  AD  as  radius  in  F.  Then  DF 
is  the  diagonal  of  a  square  equal  to  the  rectangle. 

8.  In  the  Fig.  of  183°,         CE^^sAB  .  CB, 

CD2  =  CE2  +  3ED2  =  3AB(AB  +  CB). 

9.  Show  that  the  construction  of  183°  solves  the  problem, 

"To  divide  a  segment  so  that  the  rectangle  on  the  parts 
is  equal  to  the  difference  of  the  squares  on  the  parts." 

10.  Show  that  the  construction  of  183°  solves  the  problem, 

"  To  divide  a  given  segment  so  that  the  rectangle  on 
the  whole  and  one  of  the  parts  is  equal  to  the  rectangle 
on  the  other  part  and  the  segment  which  is  the  sum  of 
the  whole  and  the  first  part." 

11.  Construct  an  equilateral  triangle  when  the  sum  of  its  side 

and  altitude  are  given.  What  does  the  double  solution 
mean?    (See  183°,  Ex.  2.) 

12.  Describe  a  square  in  a  given  acute-angled  triangle,  so 

that  one  side  of  the  square  may  coincide  with  a  side  of 
the  triangle. 

13.  Within  a  given  square  to  inscribe  a  square  having  three- 

fourths  the  area  of  the  first. 

14.  Within  an  equilateral  triangle  to  inscribe  a  second  equi- 

lateral triangle  whose  area  shall  be  one -half  that  of 
the  first. 

1 5.  Produce  a  segment  AB  to  C  so  that  the  rectangle  on  the 

sum  and  difference  of  AC  and  AB  shall  be  equal  to  a 
given  square. 

K 


146  SYNTHETIC   GEOMETRY. 

16.  Draw  a  tangent  to  a  given  circle  so  that  the  triangle 

formed  by  it  and  two  fixed  tangents  may  be  (i)  a 
maximum,  (2)  a  minimum. 

17.  Draw  a  circle  to  touch  two  sides  of  a  given  square,  and 

pass  through  one  vertex.  Generalize  this  problem  and 
show  that  there  are  two  solutions. 

18.  Given  any  two  lines  at  right  angles  and  a  point,  to  find  a 

circle  to  touch  the  lines  and  pass  through  the  point. 

19.  Describe  a  circle  to  pass  through  a  given  point  and  to 

touch  a  given  line  at  a  given  point  in  the  line. 

20.  Draw  the  oblique  lines  required  to  change  a  given  square 

into  an  octagon. 

If  the  side  of  a  square  is  24,  the  side  of  the  result- 
ing octagon  is  approximately  10 ;  how  near  is  the 
approximation  ? 

21.  The  area  of  a  regular  dodecagon  is  three  times  that  of 

the  square  on  its  circumradius. 

22.  By  squeezing  in  opposite  vertices  of  a  square  it  is  trans- 

formed into  a  rhombus  of  one-half  the  area  of  the  square. 
What  are  the  lengths  of  the  diagonals  of  the  rhombus? 

23.  P,  Q,  R,  S  are  the  middle  points  of  the  sides  AB,  BC, 

CD,  and  DA  of  a  square.  Compare  the  area  of  the 
square  with  that  of  the  square  formed  by  the  joins  AQ, 
BR,  CS,  and  DP. 

24.  ABCDEFGH  is  a  regular  octagon,  and  AD  and  GE  are 

produced  to  meet  in  K.  Compare  the  area  of  the  tri- 
angle DKE  with  that  of  the  octagon. 

25.  The  rectangle  on  the  chord  of  an  arc  and  the  chord  of 

its  supplement  is  equal  to  the  rectangle  on  the  radius 
and  the  chord  of  twice  the  supplement. 

26.  At  one  vertex  of  a  triangle  a  tangent  is  drawn  to  its  cir- 

cumcircle.  Then  the  square  on  the  altitude  from  that 
vertex  is  equal  to  the  rectangle  on  the  perpendiculars 
from  the  other  vertices  to  the  tangent. 

27.  SOT  is  a  centre-line  and  AT  a  tangent  to  a  circle  at  the 

point  A.     Determine  the  angle  AOT  so  that  AS  =  AT. 


PART    III. 


PRELIMINARY. 

1 88°.  By  superposition  we  ascertain  the  equality  or  in- 
equality of  two  given  line-segments.  But  in  order  to  express 
the  relation  between  the  lengths  of  two  unequal  segments  we 
endeavour  to  find  two  numerical  quantities  which  hold  to  one 
another  the  same  relations  in  magnitude  that  the  given  seg- 
ments do. 

Let  AB  and  CD  be  two  given  segments.  If  they  are  com- 
mensurable (150°,  5)  some  M.l.  can  be  found  with  respect  to 
which  the  measures  of  AB  and  CD  (150°,  2)  are  both  whole 
numbers.  Let  in  denote  the  measure  of  AB  and  n  the 
measure  of  CD  with  respect  to  this  unit-length. 

The  numbers  in  and  n  hold  to  one  another  the  same  rela- 
tions as  to  magnitude  that  the  segments  AB  and  CD  do. 

The  fraction  —  is  called  in  Arithmetic  or  Algebra  the  ratio 
n 

ofm  to  11,  and  in  Geometry  it  is  called  the  ratio  of  AB  to  CD. 
Now  n  has  to  in  the  same  ratio  as  unity  has  to  the  fraction 

— .     But  if  CD  be  taken  as  i/J.  its  measure  becomes  unity, 
n  ^ 

in 
while  that  of  AB  becomes  — • 

Therefore  the  ratio  of  AB  to  CD  is  the  measure  of  AB  with 
respect  to  CD  as  unit-length. 

When  AB  and  CD  are  commensurable  this  ratio  is  expres- 
sible arithmetically  either  as  a  whole  number  or  as  a  fraction  ; 
147 


148  SYNTHETIC   GEOMETRY. 

but  when  the  segments  are  incommensurable  the  ratio  can 
only  be  symbolized,  and  cannot  be  expressed  arithmetically 
except  approximately. 

189°.  If  we  suppose  CD  to  be  capable  of  being  stretched 
until  it  becomes  equal  in  length  to  AB,  the  numerical  factor 
which  expresses  or  denotes  the  amount  of  stretching  neces- 
sary may  conveniently  be  called  the  tensor  of  AB  with 
respect  to  CD.  (Hamilton.) 

As  far  as  two  segments  are  concerned,  the  tensor,  as  a 
numerical  quantity,  is  identical  with  the  ratio  of  the  segments, 
but  it  introduces  a  different  idea.  Hence  in  the  case  of  com- 
mensurable segments  the  tensor  is  arithmetically  expressible, 
but  in  the  case  of  incommensurable  ones  the  tensor  may  be 
symbolically  denoted,  but  cannot  be  numerically  expressed 
except  approximately. 

Thus  if  AB  is  the  diagonal  of  a  square  of  which  CD  is  the 
side,  AB  =  CDV2  (180°);  and  the  tensor  of  AB  on  CD, 
Le.^  the  measure  of  AB  with  CD  as  unit-length,  is  that 
numerical  quantity  which  is  symbolized  by  sji^  and  which 
can  be  expressed  to  any  required  degree  of  approximation  by 
that  arithmetical  process  known  as  "extracting  the  square 
root  of  2." 

190°.  That  the  tensor  symbolized  by  ^2  cannot  be  ex- 
pressed arithmetically  is  readily  shown  as  follows  : — 

If  /^2  can  be  expressed  numerically  it  can  be  expressed  as 

a  fraction,  — ,  which  is  in  its  lowest  terms,  and  where  accord- 
n 

ingly  m  and  n  are  not  both  even. 

If  possible  then  let  ^2  =  -. 
n 

Then  211^  =  m'^.     Therefore  m'^  and  in  are  both  even  and  « 

is  odd.  „ 

m 
But  if  m  is  even,  —  is  even,  and  n^  and  ;/  are  both  even. 

But  n  cannot  be  both  odd  and  even. 

Therefore  sji  cannot  be  arithmetically  expressed. 


PRELIMINARY.  I49 

Illustration  of  an  ificommensurable  tensor. 
Let   BD   be  equal  to  AB,  and  let  AC  be  equal  to  the 
diagonal  of  a  square  of  which  AB  is  the  side. 

^ E'F' ^ 

A  B  E   'c'   1^  D 


Then  some  tensor  will  bring  AB  to  AC. 

Let  BD  be  divided  into  10  equal  parts  whereof  E  and  F 
are  those  numbered  4  and  5. 

Then  the  tensor  i .  4  stretches  AB  to  AE,  and  tensor  i .  5 
stretches  AB  to  AF.  But  the  first  of  these  is  too  small  and 
the  second  too  great,  and  C  lies  between  E  and  F. 

Now,  let  EF  be  divided  into  10  equal  parts  whereof  E',  F' 
are  those  numbered  i  and  2. 

Then,  tensor  i .  41  brings  AB  to  AE',  and  tensor  i .  42 
brings  AB  to  AF' ;  the  first  being  too  small  and  the  second 
too  great. 

Similarly  by  dividing  E'F'  into  10  equal  parts  we  obtain 
two  points  ^,  yj  numbered  4  and  5,  which  lie  upon  opposite 
sides  of  C  and  adjacent  to  it. 

Thus,  however  far  this  process  be  carried,  C  will  always  lie 
between  two  adjacent  ones  of  the  points  last  obtained. 

But  as  every  new  division  gives  interspaces  one-tenth  of 
the  length  of  the  former  ones,  we  may  obtain  a  point  of 
division  lying  as  near  C  as  we  please. 

Now  if  AB  be  increased  in  length  from  AB  to  AD  it  must 
at  some  period  of  its  increase  be  equal  to  AC. 

Therefore  the  tensor  which  brings  AB  to  AC  is  a  real 
tensor  which  is  inexpressible,  except  approximately,  by  the 
symbols  of  Arithmetic. 

The  preceding  illustrates  the  difference  between  magnitude 
and  number.  The  segment  AB  in  changing  to  AD  passes 
through  every  intermediate  length.  But  the  commensurable 
or  numerically  expressible  quantities  lying  between  i  and  2 
must  proceed  by  some  unit  however  small,  and  are  therefore 
not  continuous. 


ISO  SYNTHETIC   GEOMETRY. 

Hence  a  mag7titude  is  a  variable  which^  in  passing  from 
one  value  to  another^  passes  through  every  intermediate  value. 

191°.  The  tensor  of  the  segment  AB  with  respect  to  AC,  or 
the  tensor  of  AB  on  AC  is  the  numerical  factor  which  brings 
AC  to  AB. 

But  according  to  the  operative  principles  of  Algebra, 

^^.AC=AB, 

AB 
.*.  -^  is  the  tensor  which  brings  AC  to  AB. 

Hence  the  algebraic  form  of  a  fraction^  when  the  parts 
denote  segments^  is  interpreted  geometrically  by  the  tensor 
which  brings  the  denominator  to  the  numerator ;  or  as  the 
ratio  of  the  numerator  to  the  denominator. 


SECTION   I. 

PROPORTION   AMONGST   LINE-SEGMENTS. 

192°.  Def — Four  line-segments  takett  in  order  form  a 
proportion^  or  are  in  proportion^  when  the  tensor  of  the  first 
on  the  second  is  the  same  as  the  tensor  of  the  third  on  the 
fourth. 

This  definition  gives  the  relation 

-r-, (^> 

where  a,  b^  c,  and  d  denote  the  segments  taken  in  order. 

The  fractions  expressing  the  proportion  are  subject  to  all 
the  transformations  of  algebraic  fractions  (158°),  and  the  re- 
sult is  geometrically  true  whenever  it  admits  of  a  geometric 
interpretation. 

The  statement  of  the  proportion  is  also  written 

(B) 


PROPORTION    AMONGST   LINE-SEGMENTS.    151 

where  the  sign  :  indicates  the  division  of  the  quantity  de- 
noted by  the  preceding  symbol  by  the  quantity  denoted  by 
the  following  symbol. 

In  either  form  the  proportion  is  read 

"  rt:  is  to  ^  as  ^  is  to  dP 

193°.  In  the  form  (b)  a  and  d  are  called  the  extremes,  and 
b  and  c  the  means ;  and  in  both  forms  a  and  c  are  called 
antecedents  and  b  and  d  consequents. 

In  the  form  (a)  a  and  d^  as  also  b  and  c^  stand  opposite 
each  other  when  written  in  a  cross,  as 
a\c 
b\d' 
and  we  shall  accordingly  call  them  the  opposites   of  the 
proportion. 

194°.  I.  From  form  (a)  we  obtain  by  cross-multiplication 
ad— be, 
which  states  geometrically  that 

When  four  segments  are  in  proportion  the  rectangle 
upon  one  pair  of  opposites  is  equal  to  that  upon  the  other 
pair  of  opposites. 

Conversely,  let  rt(5  and^i'^^'be  equal  rectangles  having  for 
adjacent  sides  a,  b,  and  a',  b'  respectively.     Then 

ab  =  a'b', 
and  this  equality  can  be  expressed  under  any  one  of  the  fol- 
lowing forms,  or  may  be  derived  from  any  one  of  them,  viz.  ; 
a  _b'     a  _a'     b  _b'     b  _a' 
a'     b     b'     b     a'     a     b'     a 
in  all  of  which  the  opposites  remain  the  same.     Therefore 

2.  Two  equal  rectangles  have  their  sides  in  proportion, 
a  pair  of  opposites  of  the  proportion  coming  from  the  same 
rectangle. 

3.  A  given  proportion  amongst  four  segments  may  be 
written  in  any  order  of  sequence,  provided  the  opposites 
remain  the  same. 


^S2  SYNTHETIC   GEOMETRY. 

195°.  The  following  transformations  are  important. 
Let  ^  =  ^,  then 

a±b    c±d  /        ,r  ■     \ 

1.  ~~b~^~T^  {a>btoY  -  sign) 

a     c    a-\-c     a  —  c         1    ^     r  •     \ 

2.  -/,  =  :>==T-r:>=7--7-       {ci>ciox  -  sign) 
b     d    b  +  a    b-d 


^       a     c 
Let  ,=-^= 
b    d 

3- 

c 

=  etc.,  then 

a     c 
b    d 

f    b  +  d+f+Qic' 

To  prove  i. 

*•■ 

a     c 
I'd' 

...|±.=f±:,and4^  = 

c±d 
d  ' 

To  prove  2. 

or 

a     c 
b~d' 

,    a     b 
'  '  "c'd' 

a±c    b±d 

c          d   ' 

a     c    a±c 

b    d    b±d' 

(194°,  3) 

To  prove  3. 

•.• 

n  __e  _ 

a  +  c     P 
-irr-^—r.^  say, 

b  7  7+Q~^+^+/ 


SIMILAR  TRIANGLES. 


196°.  Def. — I.  Two  triangles  are  similar  when  the  angles  of 
the  one  are  respectively  equal  to  the  angles  of  the  other. 

{n\  4) 

2.  The  sides  opposite  equal  angles  in  the  two  triangles  are 
corresponding  or  homologous  sides. 

The  symbol  ^  will  be  employed  to  denote  similarity,  and 
will  be  read  "is  similar  to." 


PROPORTION   AMONGST   LINE-SEGMENTS.    1 53 

In  the  triangles  ABC  and  A'B'C,  if  zA=lA'  and  z.B  =  ^B', 
then  also   lC=lC  and    the    tri- 
angles are  similar. 

The   sides    AB    and    A'B'    are         /  /     \  //> 

homologous,  so  also  are  the  other       /   /      \  // 

pairs     of     sides     opposite    equal     ^      '         ^        '^—^ 
angles.  ad  c 

Let  BD  through  B  and  B'D'  through  B'  make  the 
^BDA  =  ^B'D'A'. 

Then  AABD  «  AA'B'D'  since  their  angles  are  respectively 
equal.  In  like  manner  ADBC  «  AD'B 'C,  and  BD  and  B'D' 
divide  the  triangles  similarly. 

3.  Lines  which  divide  similar  triangles  similarly  are 
homologous  lines  of  the  triangles,  and  the  intersections  of 
homologous  /i?tes  are  homologous  points. 

Cor.  Evidently  the  perpendiculars  upon  homologous  sides 
of  similar  triangles  are  homologous  lines.  So  also  are  the 
medians  to  homologous  sides ;  so  also  the  bisectors  of  equal 
angles  in  similar  triangles  ;  etc. 

197°.  Theorem. — The  homologous  sides  of  similar  triangles 
are  proportional. 

AABC^AA'B'C 
having  lA.  =  lA.' 

and  lR=lB'. 

rj,.^  AB      BC      CA 

^^'^  A^=FC  =  C^-         , 

B  C 

Proof.— V\3,ce  A'  on  A,  and  let  C  fall  at  D.  Then,  since 
lA'=lA,  A'B'  will  lie  along  AC  and  B'  will  fall  at  some  point 
E.  Now,  AA'B'C'^AAED,  and  therefore  ^AED=^B, 
and  B,  D,  E,  C  are  concyclic.  (107°) 

Hence  AD.AB  =  AE.AC,  (176°,  2) 

or  A'C'.AB  =  A'B'.AC.  , 

AB      AC  /ox 

A^B'=A^-  (^94,2) 


154  SYNTHETIC   GEOMETRY. 

Similarly,  by  placing  B'  at  B,  we  prove  that 
AB  _BC 
A'B'     B'C* 

AB^BC^CA  , 

A'B'     B'C     C'A''  ^'  '  ' 

Cor.  I.  Denoting  the  sides  of  ABC  by  a,  b,  c,  and  those  of 


A'B'C  by  a',  b',  c', 


a  _b  _c 
a~b'~c'' 


^  a     b     c       a  +  b  +  c  /,^r-°   ^\ 

Cor.  2.  v  =  A'  =  7'  =  vn:i'3:7>'  ('95  ,  3) 

a     0     c     a  +b  +c 

i.e.,  the  perimeters  of  similar  triangles  are  proportional  to 

any  pair  of  homologous  sides. 


198°.  Theorem. — Two  triangles  which  have  their  sides  pro- 
b'        portional   are   similar,   and  have 
their  equal  angles  opposite  hom- 
ologous sides.    (Converse  of  197°.) 
AB  _  BC  _  CA 
A'B'     B'C     C'A'' 
Then    .iA  =  zj\',     ^B  =  z.B',     and 
^C=z.C'. 

Proof.— On  A'C  let  the  AA'DC  be  constructed  so  as  to 
have  the  ^DA'C'=zA 

and  ^DC'A'=Z-C. 

Then  AA'DC ^AABC,  (196°,  Def.  i) 

^^A  AB      AC      BC  .  ,.s 

^^^  A^==A^'  =  DC'  ^^97) 

u  .  AB      AC      BC  /,       X 

^"'  A^  =  A^'  =  B'C'  ^^^P-^ 

A'D  =  A'B' 
and  DC'  =  B'C, 

and  AA'DC  =  AA'B'C'.  (58°) 

ZJ\'  =  ^A,  ^B'=^B, 
and  ^C'=^C.  q.e.d. 


PROPORTION   AMONGST   LINE-SEGMENTS.    1 55 

199°.  Theorem. — If  two  triangles  have  two  sides  in  each 
proportional  and   the  included  a 

angles  equal,  the  triangles  are 
similar.  D^ 

then      AABC^AA'B'C. 

c 

Proof. — Place  A'  on  A,  and  let  A'C  lie  along  AB,  and 

A'B'  lie  along  AC,  so  that  C  falls  at  D  and  B'  at  E. 

The  triangles  AED  and  A'B'C  are  congruent  and  therefore 

.    .,  ,  AB     AC 

similar,  and  ^=^ 

Hence                   AB  .  AD  =  AE .  AC  ;  ( 1 94°) 

and  .'.  B,  D,  E,  C  are  concyclic.  {mT) 

Z_AED  =  aB,  and  £ADE  =  zlC,  (106°,  Cor.  3) 

and                   AABC  ^  AAED  «  AA'B'C.  q.e.d. 

200°.  Theorem. — If  two  triangles  have  two  sides  in  each 
proportional,  and  an  angle  opposite  a  homologous  side  in 
each  equal : 

1.  If  the  angle  is  opposite  the  longer  of  the  two  sides  the 
triangles  are  similar. 

2.  If  the  angle  is  opposite  the  shorter  of  the  two  sides  the 
triangles  may  or  may  not  be 
similar. 

I.  If         BOAB, 

AABC^AA'B'C. 

Pr^^/— Place  A'  at  A  and  let  B'  fall  at  D,  and  A'C  along 
AC.     Draw  DE  ||  to  BC.     Then 

AABC«  AADE,  and  ^^=§§- 


156  SYNTHETIC  GEOMETRY. 

And  since  B'C'>  A'B',  the  AA'B'C'  =  AADE  and  they  are 
therefore  similar.  (65°,  i) 

But  AABC^AADE, 

AABC^AA'B'C. 

2.  If  BC<AB,  B'C'<A'B',  and  the  triangles  may  or  may 
not  be  similar. 

Proof.— Since  AD  =  A'B',  and  DE  =  B'C',  and  B'C'<A'B', 
.*.  the  triangles  A'B'C  and  ADE  may  or  may  not  be 
congruent  (65°,  2),  and  therefore  may  or  may  not  be 
similar. 

But  AABC^AADE, 

.*.  the    triangles    ABC    and    A'B'C    may  or  may   not  be 
similar. 

Cor.  Evidently,  if  in  addition  to  the  conditions  of  the 
theorem,  the  angles  C  and  C  are  both  less,  equal  to,  or 
greater  than  a  right  angle  the  triangles  are  similar. 

Also,  if  the  triangles  are  right-angled  they  are  similar. 

201°.  The  conditions  of  similarity  of  triangles  may  be 
classified  as  follows  : — 

1.  Three  angles  respectively  equal.     (Def.  of  similarity.) 

2.  Three  sides  proportional. 

3.  Two  sides  proportional  and  the  included  angles  equal. 

4.  Two  sides  proportional  and   the  angles   opposite  the 

longer  of  the  homologous  sides  in  each  equal. 

If  in  4  the  equal  angles  are  opposite  the  shorter  sides  in 
each  the  triangles  are  not  necessarily  similar  unless  some 
other  condition  is  satisfied. 

By  comparing  this  article  with  66°  we  notice  that  there  is  a 
manifest  relation  between  the  conditions  of  congruence  and 
those  of  similarity. 

Thus,  if  in  2,  3,  and  4  of  this  article  the  words  "propor- 
tional "  and  "  homologous "  be  changed  to  "  equal,"  the 
statements  become  equivalent  to  1,2,  and  5  of  Art.  66°.     The 


PROPORTION   AMONGST   LINE-SEGMENTS.    1 5/ 

difference  between  congruence  and   similarity  is  the  non- 
necessity of  equality  of  areas  in  the  latter  case. 

When  two  triangles,  or  other  figures,  are  similar,  they  are 
copies  of  one  another,  and  the  smaller  may  be  brought,  by  a 
uniform  stretching  of  all  its  parts,  into  congruence  with  the 
larger.  Thus  the  primary  idea  of  similarity  is  that  every 
line-segment  of  the  smaller  of  two  similar  figures  is  stretched 
to  the  same  relative  extent  to  form  the  corresponding  seg- 
ments of  the  larger  figure.  This  means  that  the  tensors  of 
every  pair  of  corresponding  line- segments,  one  from  each 
figure,  are  equal,  and  hence  that  any  two  or  more  line- 
segments  from  one  figure  are  proportional  to  the  correspond- 
ing segments'  from  the  second  figure. 

Def. — Two  line-segments  are  divided  similarly  when,  being 
divided  into  the  same  number  of  parts,  any  two  parts  from 
one  of  the  segments  and  the  corresponding  parts  from  the 
other  taken  in  the  same  order  are  in  proportion. 

202°.  Theorejn. — A  line  parallel  to  the  base  of  a  triangle 
divides  the  sides  similarly  ;  and 

Conversely,  a  line  which  divides  two  sides  of  a  triangle 
similarly  is  parallel  to  the  third  side.  B 

DE   is   II  to  AC.     Then  BA  and  BC  are 
divided  similarly  in  D  and  E. 

Proof.— Tht  triangles  ABC  and  DBE  are    /^ 
evidently  similar,  ^  ^ 

AB     CB  ^.    .    AD     CE  .  ^_o     ^ 

DB  =  E-B'^"^-  DB^EB'  ^^^^  '  ^^ 

and  AB  and  CB  are  divided  similarly  in  D  and  E.  g.e.d. 

Conversely,  if  DE  so  divides  BA  and  BC  that 
AD  :  DB  =  CE  :  EB,  DE  is  ||  to  AC. 
P.../-Since  AD^CE     ._  AB^CB  ^   ^„,  ^,^  .^j.^g,,, 

ABC  and  DBE  having  the  angle  B  common,  and  the  sides 


158 


SYNTHETIC  GEOMETRY. 


about  that  angle  proportional,  are  similar.  (i99°) 

aBDE=^  and  DE  is||  to  AC.  q.e.d. 

Cor.  I.  Since  the  triangles  ABC  and  DBE  are  similar 
BA:BD=AC:DE. 


203°.  Theorem. — Two  transversals  to  a  system  of  parallels 
are  divided  similarly  by  the  parallels. 

AA'  is  II  to  BB'  is  ||  to  CC,  etc. 
Then  AD  and  A'D'  are  divided  similarly. 

Proof. — Consider  three  of  the  ||s,  AA', 
BB',  and  CC,  and  draw  A'Q  ||  to  AD. 

Then    AP    and    BQ    are    1 — 7s,   and 

AB  =  A'P  and  BC  =  PQ.     (81°,  i) 

But  A'QC  is  a  triangle  and  PB'  is  ||  to  QC. 

A'P     A'B'  .  ^,0.   ^,.  AB      BC    .       o     V 
^=—  (202),  01  ^,g,=^^,  (194,3). 


Similarly,  if  DD'  be  a  fourth  parallel, 


AB 
A'B' 


BC 
B'C 


CD 
CD' 


BC^ 

B'C 


CD 

CD" 


=  etc. 


Def, — A  set  of  three  or  more  lines  meeting  in  a  point  is  a 
pencil  and  the  lines  are  rays. 

The  point  is  the  vertex  or  centre  of  the  pencil. 

Cor.  I.  Let  the  transversals  meet  in  O,  and  let  L  denote 
any  other  transversal  through  O. 

Then    AD,  A'D',   and    L    are   all    divided    similarly    by 

the  parallels.  But  the  parallels  are  transversals  to  the 
pencil. 

.*.  parallel  transversals  divide  the  rays  of  a  pencil 
similarly. 

Cor.  2.  Applying  Cor.  i  of  202", 

OA     OB     OC     OD 


AA'     BB'     CC     DD" 


=etc. 


PROPORTION   AMONGST   LINE-SEGMENTS.    1 59 


204°.   Theorem. — The  rectangle  on  any  two  sides  of  a  tri- 
angle is  equal  to  twice  the  rectangle  on  B 

the  circumradius  (97°,  Def.)  and  the  alti- 
tude to  the  third  side.  ^f^ ^-^ — ^° 

BD  is  _L  to  AC  and  BE  is  a  diameter. 
Then  BA.BC  =  BE.BD. 


Proof.—  lA  =  lE, 

and^ADB=z.ECB=~l, 


(106°,  Cor.  I) 
(106°,  Cor.  4) 

AABD^AEBC,  and  |^=|§, 
BA.BC  =  BE.BD.  g.e.d. 

Cor.  Denoting  BD  by  p  and  the  circumradius  by  R, 
ac=  2/R, 
and  multiplying  by  3,  and  remembering  that  pd=2^  (i  53°,  2), 

4A' 

which  (with  i7si°,  Ex.  i)  gives  the  means  of  calculating  the 
circumradius  of  a  triangle  when  its  three  sides  are  given . 


we  obtain 


205°.  Theorem. — In  a  concyclic  quadrangle  the  rectangle 
on  the  diagonals  is  equal  to  the  sum  of  the  rectangles  on  the 
sides  taken  in  opposite  pairs. 

AC .  BD  =  AB  .  CD  +  BC .  AD. 

Proof.— DvdiVf     AE     making     aAED 
=  zABC.     Then,   since  ^BCA=^BDA, 
the  triangles  EDA  and  BCA  are  similar. 
BC.AD=AC.DE. 

Again,  since  z_AEB  is  supp.  to  Zj\ED, 
and  Z.CDA  is  supp.  to  ^lABC,  therefore  triangles  BE  A  and 
CD  A  are  similar,  and  AB  .  CD  =  AC .  EB. 

Adding  these  results,  AB .  CD  +  BC .  AD  =  AC .  BD. 

This  theorem  is  known  as  Ptolemy's  Theorem. 


206°.  Def. — Two  rectilinear  figures  are  similar  when  they 


l60  SYNTHETIC   GEOMETRY. 

can  be  divided  into  the  same  number  of  triangles  similar  in 

pairs  and  similarly  placed. 
Thus  the  pentagons  X- 
and  Y  can  be  divided  into 
the  same  number  of  tri- 
angles. 

If  then  AP  «  AP', 
AQ  ^  AQ',  AR  «  AR', 
and  the  triangles  are  similarly  placed,  the  pentagons  are 
similar. 

The  triangles  are  similarly  placed  if  Z.EAD  corresponds  to 
E'A'D',  ^AED  to  A'E'D',  .lDAC  to  D'A'C,  etc. 

This  requires  that  the  angles  A,  B,  C,  etc.,  of  one  figure 
shall  be  respectively  equal  to  the  angles  A',  B',  C,  etc.,  of  the 
other  figure. 

Hence  when  two  rectilinear  figures  are  similar,  their  angles 
taken  in  the  same  order  are  respectively  equal,  and  the  sides 
about  equal  angles  taken  in  the  same  order  are  pro- 
portional. 

Line-segments,  such  as  AD  and  A'D',  which  hold  similar 
relations  to  the  two  figures  are  similar  or  homologous  lines  of 
the  figures. 

207°.  Theorem. — Two  similar  rectilinear  figures  have  any 
two  line-segments  from  the  one  proportional  to  the  homolo- 
gous segments  from  the  other. 

Proof. — By  definition  AP^^AP',  and  they  are  similarly 
placed,  .-.  AE  :  A'E'  =  AC  :  A'C. 

For  like  reasons,   AD  :  A'D'  =  AC  :  A'C'  =  AB  :  A'B'. 
AE  _  AD  _  AC  _  AB  _ 
A'E'    A'D'     A'C     A'B'         '' 
and  the  same  can  be  shown  for  any  other  sets  of  homologous 
line-segments. 

Cor.  I.  All  regular  polygons  of  the  same  species  are  similar 
figures. 


PROPORTION   AMONGST   LINE-SEGMENTS.    l6l 

Now,  let  a^  b,  c,  ...,  a',  b\  (f, ...,  be  homologous  sides  of  two 
similar  regular  polygons,  and  let  r  and  r'  be  their  circumradii. 
Then  r  and  r'  are  homologous, 

r _a  _b _c _      _  a-^b-^c^...         /       o     . 
V~a'~'b~'<f~""'V+Y+l^\::       ^'^^'^^ 
_  perimeter  of  P 
perimeter  of  P' 
But  at  the  limit  (148°)  the  polygon  becomes  its  circumcircle. 

.•.  the  circumferences  of  any  two  circles  are  proportional  to 
their  radii. 

Cor.  2.  If  ^,  c'  denote  the  circumferences  of  two  circles  and 

r  and  r'  their  radii,      -  =  ^  =  constant. 
r    r 

Denote  this  constant  by  27r,  then 

C=7.Trr. 

It  is  shown  by  processes  beyond  the  scope  of  this  work 
that  TT  stands  for  an  incommensurable  numerical  quantity, 
the  approximate  value  of  which  is  3. 141 5926... 

Cor.  3.  Since  equal  arcs  subtend  equal  angles  at  the  centre 
(102°,  Cor.  2),  if  s  denotes  the  length  of  any  arc  of  a  circle 
whose  radius  is  r,  the  tensor  -  varies  directly  as  s  varies, 
and  also  varies  directly  as  the  angle  at  the  centre  varies. 

Hence  -  is  taken  as  the  measure  of  the  angle,  subtended 
by  the  arc,  at  the  centre.     Denote  this  angle  by  Q.     Then 

and  when  s  =  r^B  becomes  the  unit  angle. 

.*.  the  unit  angle  is  the  angle  subtended  at  the  centre  by  an 
arc  equal  in  length  to  the  radius. 

This  unit  is  called  a  radian,  and  the  measure  of  an  angle 
in  radians  is  called  its  radian  measure.  Radian  measure  will 
be  indicated  by  the  mark  '^. 

Cor.  4.  When  j=^  =  a  semicircle,  Q  —  ir. 


1 62  SYNTHETIC   GEOMETRY. 

But  a  semicircle  subtends  a  straight  angle  at  the  centre. 
.*.  tr  is  the  radian  measure  of  a  straight  angle  and  -  of  a     J. 

Now  a  straight  angle  contains  i8o°,  (4i°) 

7r'^=i8o°. 
Hence  i'^=57°.29578..., 

and  i°  =  o'\oi7453...  ; 

and  these  multipliers  serve  to  change  the  expression  of  a 
given  angle  from  radians  to  degrees  or  from  degrees  to 
radians. 

Cor.  5.  Since  the  area  of  a  circle  is  equal  to  one-half  that 

of  the  rectangle  on  its  radius  and  a  segment  equal  in  length 

to  its  circumference,  (i49°) 

Q)  =  \cr=\.2Trr.r  (Cor.  2) 

=  7rr2. 

.'.  the  area  of  a  0  is  tt  times  that  of  the  square  on  its  radius. 

208°.   Theorem. — The  bisectors  of  the  vertical  angle  of  a 
triangle  each  divides  the  base  into  parts  which  are  propor- 
g  tional  to  the  conterminous 

sides. 

BD    and    BD'    are    bi- 
sectors of  Z.B.     Then 
AD_AD^_AB 
DC     CD'     BC* 

/*r^^— Through  C  draw 
EE'lltoAB.    Then 
EBE'='~1(45°),  and^E  =  ^ABD  =  ^DBC, 

BC  =  EC  =  CE'.  (88°,  3) 

But  ABD  and  ABD'  are  triangles  having  EE'  ||  to  the 
common  base  AB. 

AB_AD      ^  AB     AD'        ,      o  ^     v 

EC-D-C  ^"^  CE'  =  CD"       ^^°^  '  ^^'-^ 

^_  AD_AD'    AB  , 

DC~Ciy  =  BC-  ^^'^'^• 

Cor.  D  and  D'  divide  the  base  internally  and  externally  in 


PROPORTION   AMONGST   LINE-SEGMENTS.    163 

the  same  manner.     Such  division  of  a  segment  is  called  har- 
monic division. 

.*.  the  bisectors  of  any  angle  of  a  triangle  divide  the  oppo- 
site side  harmonically. 

209°.  Theorem.— K  line  through  the  vertex  of  a  triangle 
dividing  the  base  into  parts  which  are  proportional  to 
the  conterminous  sides  is  a  bisector  of  the  vertical  angle. 
(Converse  of  208°.) 

Let  the  line  through  B  cut  AC  internally  in  F.     Then,  AD 

being  the  internal  bisector  |^=^^  (208°),  and  ||  =  ^-^  by 

,        ,,     .      .  AF     AD 

hypothesis,..  FC^DC- 

But  AD  is  <  AF  while  DC  is  >  FC. 

.'.  the  relation  is  impossible  unless  F  and  D  coincide,  i.e.^ 
the  line  is  the  bisector  AD. 

Similarly  it  may  be  proved  that  if  the  line  divides  the  base 
externally  it  is  the  bisector  AD'. 


210°.  Theorem. — The  tangent  at  any  point  on  a  circle  and 
the  pei*pendicular  from  that  point  upon  the  diameter  divide 
the  diameter  harmonically. 

AB  is   divided  harmonically  in 
M  and  T.  a 

Proof.-LQ,VT=^LVWY=~^,  (110°) 
ACPM^ACTP, 


r 


and 


CM^CP 

CP     CT' 

CB+CM     CT+CB 


or 


CM_CB 
CB  CT 
AM     AT 


CB-CM     Ct-CB'  MB     BT' 

,\  AB  is  divided  harmonically  in  M  and  T. 


(195°,  2) 
q.e.d. 


211".  The  following  examples  give  important  results. 


164 


SYNTHETIC   GEOMETRY. 


Ex.  I.  L,  M,  and  N  are  tangents  which  touch  the  circle  at 

A,  B,  and  P. 

AX  and  BY  are  _Ls  on 
N,  PC  is  X  on  AB,  and 
PO  and  PR  are  ±s  upon 
L  and  M. 

Let  N  meet  the  chord 

of  contact  of  L  and  M  in 

T.     Then   the  triangles 

TAX,  TPC,  TBY  are  all 

TA.TB_TP2 

PC2' 

(176°,  Cor.  2) 
(A) 


similar, 
But 


AX.  BY 


Then 


(114°,  Cor.  i) 


1^ 


and 

and 

Similarly 
Ex. 


TA_TP_TB. 

AX     PC     BY ' 

TA.TB  =  TP2, 

AX.BY  =  PC2 

Again,  let  L  and  N  intersect  in  V. 
VP=VA, 

/.V0P  =  ^VXA  =  ~1, 
^QVP  =  z.XVA. 
AVXA  =  AVQP, 
AX  =  PQ. 

BY  =  PR,   .-.   PO.PR  =  PC^ (B) 

AD  is  a  centre-line  and  DO  a  perpendicular  to  it, 
and  AQ  is  any  line  from  A  to 
the  line  DO. 

Let  AQ  cut  the-  circle  in  P. 
Then       AADQ^AAPB, 
AD^AQ 
AP     AB' 
or  AD.AB  =  AP.AO. 

But  the  circle  and  the  point  D 
being  given,  AB .  AD  is  a  given 
constant. 

AP  .  AQ  =  a  constant. 
Conversely,  if  Q  moves  so  that  the  cziAP.AQ  remains 
constant,   the    locus  of  Q  is   a  line  ±  to  the  centre-line 
through  A.        , 


PROPORTION  AMONGST  LINE-SEGMENTS.      165 

Now,  let  the  dotted  lines  represent  rigid  rods  of  wood  or 
metal  jointed  together  so  as  to  admit  of  free  rotation  about 
the  points  A,  C,  P,  U,  V,  and  Q,  and  such  that  UPVQ  is  a 
rhombus  (82°,  Def.  i),  and  AU=AV,  and  AC  =  CP,  AC 
being  fixed. 

PQ  is  the  right  bisector  of  UV,  and  A  is  equidistant  from 
U  and  V.     Therefore  A,  P,  Q  are  always  in  line. 

Also,  PUO  is  an  isosceles  triangle  and  UA  is  a  line  to  the 
base,  therefore  UA2-UP2  =  AP .  AQ  (174°).,.  But,  UA  and 
UP  being  constants,  AP  .  AQ  is  constantl"  ' ''' 

And  AC  being  fixed,  and  CP  being  equal  to  AC,  P  moves 
on  the  circle  through  A  having  C  as  centre. 
.'.  Q  describes  a  line  _L  to  AC. 

This  combination  is  known  as  Peaucelller's  cellj  and  is 
interesting  as  being  the  first  successful  attempt  to  describe  a 
line  by  circular  motions  only. 

Ex.  3.  To  construct  an  isosceles  triangle  of  which  each 
basal  angle  shall  be  double  the  vertical  angle. 

Let  ABC  be  the  triangle  required,  and  let  AD 
bisect  the  Z.A. 

Then  ^B=z_BAD=^DAC,  and  ^C  is  common 
to  the  triangles  ABC  and  DAC.  Therefore  these 
triangles  are  similar,  and  the  ACAD  is  isosceles 
and  AD  =  AC. 

Also,  AABD  is  isosceles  and  AD  =  DB=AC. 
BA:AC  =  AC:DC, 
or  BC:BD  =  BD:DC. 

BC.DC  =  BD2. 
And  BC  is  divided  into  extreme  and  mean  ratio  at  D  (183°, 
Ex.  i).     Thence  the  construction  is  readily  obtained. 

Cor.  I.  The  isosceles  triangle  ADB  has  each  of  its  basal 
angles  equal  to  one-thtrd  its  vertical  angle. 

Cor.  2.  JJ\BC  =  36°,  ^BAC  =  72°,  Z-BDA=  108°.     Hence 
(i)  Ten  triangles  congruent  with  ABC,  placed  side  by  side 
with  their  vertices  at  B,  form  a  regular  decagon.  (132°) 


1 66 


SYNTHETIC   GEOMETRY. 


(2)  The  bisectors  AD  and  CE  of  the  basal  angles  of  the 

AABC  meet  its  circumcircle  in  two  points  which,  with 
the  three  vertices  of  the  triangle,  form  the  vertices  of  a 
regular  pentagon. 

(3)  The  ^BDA  =  the  internal  angle  of  a  regular  pentagon. 


The  following  Mathematical  Instruments  are  im- 
P    portant : — 

I.  Proportional  Compasses. 
This  is  an  instrument  primarily  for  the  purpose 
of  increasing  or  diminishing  given  line-segments 
in  a  given  ratio  ;  ?>.,  of  multiplying  given  line- 
segments  by  a  given  tensor. 

If  AO  =  BO  and  QO  =  PO,  the  triangles  AOB, 
POQ  are  isosceles  and  similar,  and 
AB:PQ  =  OA:OP. 
B       Hence,  if  the  lines  are  one  or  both  capable  of 
rotation  about  O,  the  distance  AB  may  be  made  to  vary  at 
pleasure,  and  PQ  will  remain  in  a  constant  ratio  to  AB. 

The  instrument  usually  consists  of  two  brass  bars  with 
slots,  exactly  alike,  and  having  the  point  of  motion  O  so 
arranged  as  to  be  capable  of  being  set  at  any  part  of  the  slot. 
The  points  A,  B,  P,  and  Q  are  of  steel. 

2.  The  Sector. 

This  is  another  instrument  which  pri- 
marily serves  the  purpose  of  increasing  or 
diminishing  given  line-segments  in  given 
ratios. 

This  instrument  consists  of  two  rules 
equal  in  length  and  jointed  at  O  so  as  to 
be  opened  and  shut  like  a  pair  of  com- 
passes.    Upon  each  rule  various  lines  are 
drawn  corresponding  in  pairs,  one  on  each  rule. 

Consider  the  pair  OA  and  OB,  called  the  "line  of  lines." 
Each  of  the  lines  of  this  pair  is  divided  into  10  equal  parts 


PROPORTION  AMONGST  LINE-SEGMENTS.      1 6/ 

which  are  again  subdivided.  Let  the  divisions  be  numbered 
from  o  to  lo  along  OA  and  OB,  and  suppose  that  the  points 
numbered  6  are  the  points  P  and  Q.  Then  GAB  and  OPQ 
are  similar  triangles,  and  therefore  PQ  :  AB=OP  :  OA.  But 
OP  =  j%AO.  .-.     PQ  =  i%AB. 

And  as  by  opening  the  instrument  AB  may  be  made  equal  to 
any  segment  not  beyond  the  compass  of  the  instrument,  we 
can  find  PQ  equal  to  ^q  of  any  such  given  segment. 

The  least  consideration  will  show  that  the  distance  5-5  is 
|AB,  3-3  is  f^^AB,  etc.  Also  that  3-3  is  f  of  7-7,  5-5  is  f  of 
7-7,  etc.  Hence  the  instrument  serves  to  divide  any  given 
segment  into  any  number  of  equal  parts,  provided  the  num- 
ber is  such  as  belongs  to  the  instrument. 

The  various  other  lines  of  the  sector  serve  other  but  very 
similar  purposes. 


3.  TAe  Pantagraph  or  Eidograph. 

Like  the  two  preceding  in- 
struments the  pantagraph  pri-  y^ 
marily  increases  or  diminishes 
segments  in  a  given  ratio,  but 
unlike  the  others  it  is  so  ar- 
ranged as  to  be  continuous  in 
its  operations,  requiring  only 
one  setting  and  no  auxiliary 
instruments. 

It  is  made  of  a  variety  of 
forms,  but  the  one  represented 
in  the  figure  is  one  of  the  most  convenient. 

AE,  AB,  and  BF  are  three  bars  jointed  at  A  and  B.  The 
bars  AE  and  BF  are  attached  to  the  wheels  A  and  B  respec- 
tively, which  are  exactly  of  the  same  diameter,  and  around 
which  goes  a  very  thin  and  fle^xible  steel  band  C. 

The  result  is  that  if  AE  and  BF  are  so  adjusted  as  to  be 
parallel,  they  remain  parallel  however  they  be  situated  with 
respect  to  AB.     E,  F  are  two  points  adjustable  on  the  bars 


6S 


SYNTHETIC    GEOMETRY. 


AE  and  BF,  and  D  is  a  point  in  line  with  EF,  around  which 
the  whole  instrument  can  be  rotated. 

Now  let  EG  KM  be  any  figure  traced  by  the  point  E  ;  then 
F  will  trace  a  similar  figure  FHLN. 

Evidently  the  triangles  DAE  and  DBF  remain  always 
similar  however  the  instrument  is  transformed.  Therefore 
DF  is  in  a  constant  ratio  to  DE,  viz.,  the  ratio  DB  :  DA. 

Now,  when  E  comes  to  G,  F  comes  to  some  point  H  in 
fine  with  GD,  and  such  that  DH  :  DG  =  DB  :  DA. 

.*.  the  triangles  EDG  and  FDH  are  similar,  and  FH  is  [[ 
to  EG,  and  has  to  it  the  constant  ratio  DB  :  DA.  Similarly 
HL  is  II  to  GK  and  has  to  it  the  same  constant  ratio,  etc. 

.'.  the  figures  are  similar,  and  the  ratio  of  homologous  lines 
in  GMandHN  is  AD  :DB. 

The  three  points  E,  D,  and  F  being  all  adjustable  the  ratio 
can  be  changed  at  pleasure. 

Altogether  the  Pantagraph  is  a  highly  important  instrument, 
and  when  so  adjusted  that  E,  D,  and  F  are  not  in  line  its 
results  offer  some  interesting  geometrical  features. 


4       : 

i        'c 

>       1 

c 

)           A 

4.  The  Diagonal  Scale. 
This  is  a  divided  scale  in 
which,  by  means  of  similar  tri- 
angles, the  difficulty  of  reading 
off  minute  divisions  is  very  much 
diminished. 

Its  simpler  form  is  illustrated  in 
°'      ^    the  figure. 
A  scale  divided  to  fortieths  of  an  inch  is,  on  account  of  the 
closeness  of  the  divisions,  very  difficult  to  read. 

In  the  scale  represented  OA  is  \  inch.  The  distance  AB 
is  divided  into  10  equal  parts  by  the  horizontal  parallel  lines 
numbered  i,  2,  3,  etc.  Then  QBO'  is  a  triangle  whereof  the 
horizontal  lines  are  all  parallel  to  the  base.  Hence  it  is 
readily  seen  from  the  proportionality  of  the  homologous  sides 
of  the  similar  triangles  formed  that  the  intercept  on  the  hori- 


PROPORTION  AMONGST  LINE-SEGMENTS.      1 69 

zontal  line  i,  between  GO'  and  OB,  is  tgO'E,  that  is 
j^  inch. 

Similarly  the  intercept  on  the  horizontal  line  2  is  ^  inch, 
on  3,  -£'q  inch,  etc. 

Hence  from  j?^  to  g^  is  one  inch  and  seven-fortieths. 

In  a  similar  manner  diagonal  scales  can  be  made  to  divide 
,any  assumed  unit-length  into  any  required  number  of  minute 
parts. 

The  chief  advantages  of  such  scales  are  that  the  minute 
divisions  are  kept  quite  distinct  and  apparent,  and  that  errors 
are  consequently  avoided. 


Exercises. 

1.  ABCD  is  a  square  and  P  is  taken  in  BC  so  that  PC  is 

one-third  of  BC.  AC  cuts  thie  diagonal  BD  in  O,  and 
AP  cuts  it  in  E.     Then  OE  is  one-tenth  of  DB. 

2.  If,  in  I,  OE  is  one-eighth  of  DB,  how  does  P  divide  BC  ? 

3.  If  BP  is  one  nth  of  BC,  what  part  of  DB  is  OE  ? 

4.  Given  three  line-segments  to  find  a  fourth,  so  that  the 

four  may  be  in  proportion. 

5.  The  rectangle  on  the  distances  of  a  point  and  its  chord  of 

contact  from  the  centre  of  a  circle  is  equal  to  the 
square  on  the  radius  of  the  circle. 

6.  OD  and  DO  are  fixed  lines  at  right  angles  and  O  is  a 

fixed  point.  A  fixed  circle  with  centre  on  OD  and 
passing  through  O  cuts  OQ  in  P.  Then  OP .  OQ  is  a 
constant  however  OQ  be  drawn. 

7.  To  divide  a  given  segment  similarly  to  a  given  divided 

segment. 

8.  To  divide  a  given  segment  into  a  given  number  of  equal 

parts. 

9.  Two  secants  through  A  cut  a  circle  in  B,  D,  and  C,  E 

respectively.     Then  the  triangles  ABE  and  ACD  are 
similar.     So  also  are  the  triangles  ABC  and  AED. 
10.  Two  chords  are  drawn  in  a  circle.     To  find  a  point  on 


lyo  SYNTHETIC   GEOMETRY. 

the  circle  from  which  perpendiculars  to  the  chords  are 
proportional  to  the  lengths  of  the  chords. 

11.  ABC  is  a  triangle  and  DE  is  parallel  to  AC,  D  being  on 

AB  and  E  on  CB.     DC  and  AE  intersect  in  O.     Then 
BO  is  a  median. 

12.  If  BO,  in  II,  cuts  DE  in  P  and  AC  in  O,  BO  is  divided 

harmonically  by  P  and  Q. 

13.  A  and  B  are  centres  of  fixed  circles  and  AX  and  BY  are 

parallel  radii.     Show  that  XY  intersects  AB   in  a 
fixed  point. 

14.  In  the  triangle  ABC,  BD  bisects  the  ^B  and  cuts  AC  in 

D.      Then    BD2=AB.  BC -AD  .  DC.     (Employ  the 
circumcircle.) 

15.  ABC  is  right-angled  at  B  and  BD  is  the  altitude  on  AC. 

(i)  The  As  ADB  and  BDC  are  each  similar  to  ABC. 
(2)  Show  by  proportion  that  AB2  =  AD  .  AC, 
and  BD2=AD.DC. 

16.  If  R  and  r  denote  the  radii  of  the  circumcircle  and  in- 

circle  of  a  triangle,  2Rr(rt  -\-d  +  c)  =  abc. 

17.  In  an  equilateral  triangle  the  square  on  the  side  is  equal 

to  six  times  the  rectangle  on  the  radii  of  the  circum- 
circle and  incircle. 

18.  OA,  OB,  OC  are  three  lines.     Draw  a  line  cutting  them 

so  that  the  segment  intercepted  between  OA  and  OC 
may  be  bisected  by  OB. 

19.  What  is  the  measure  of  an  angle  in  radians  when  its 

measure  in  degrees  is  68°  1 7'  .'* 

20.  How  many  radians  are  in  the  angle  of  an  equilateral  A  ? 

21.  The  earth's  diameter  being  7,960  miles,  what  is  the  dis- 

tance in  miles  between  two  places  having  the  same 
longitude  but  differing  16°  in  latitude.'' 

22.  Construct    a    regular   pentagon,   a    regular  decagon,   a 

regular  polygon — of  1 5  sides,  of  30  sides,  of  60  sides. 

23.  ABODE  is  a  regular  pentagon. 

(i)  Every  diagonal  is  divided  into  extreme  and  mean 
ratio  by  another  diagonal. 


PROPORTION  AMONGST  LINE-SEGMENTS.      I/I 

(2)  The  diagonals  enclose  a  second  regular  pentagon. 

24.  Compare  the  side  and  the  areas  of  the  two  pentagons  of 

23  (2). 

25.  If  one  side  of  a  right-angled  triangle  is  a  mean  propor- 

tional between  the  other  side  and  the  hypothenuse,  the 
altitude  from  the  right  angle  divides  the  hypothenuse 
into  extreme  and  mean  ratio. 

26.  A  variable  line  from  a  fixed  point  A  meets  a  fixed  circle 

in  P,  and  X  is  taken  on  AP  so  that  AP .  AX  =  a  con- 
stant.    The  locus  of  X  is  a  circle. 

27.  If  two  circles  touch  externally  their  common  tangent  is  a 

mean  proportional  between  their  diameters. 

28.  Four  points  on  a  circle  are  connected  by  three  pairs  of 

lines.  If  a,  oj  denote  the  perpendiculars  from  any  fifth 
point  on  the  circle  to  one  pair  of  lines,  /3,  ^^  to  another 
pair,  and  7,  y^  to  the  third  pair,  then  aai=/3j3i  =  77i. 
(Employ  204°.) 

29.  A  line  is  drawn  parallel  to  the  base  of  a  trapezoid  and 

bisecting  the  non-parallel  sides.  Compare  the  areas 
of  the  two  trapezoids  formed. 

30.  Draw  two  lines  parallel  to  the  base  of  a  triangle  so  as  to 

trisect  the  area. 

31.  ABC  is  right-angled  at  B,  and  AP  is  the  perpendicular 

from  A  to  the  tangent  to  the  circumcircle  at  B.  Then 
AP.AC  =  AB-. 


SECTION   II. 
FUNCTIONS  OF  ANGLES.— AREAL  RELATIONS. 

212°.  De/. — When  an  element  of  a  figure  undergoes  change 
the  figure  is  said  to  vary  that  element. 

If  a  triangle  changes  into  any  similar  triangle  it  varies  its 
magnitude  while  its  form  remains  constant ;  and  if  it  changes 


1/2  SYNTHETIC   GEOMETRY. 

into  another  form  while  retaining  the  same  area,  it  varies  its 
form  while  its  area  remains  constant,  etc. 

Similar  statements  apply  to  other  figures  as  well  as  triangles. 

When  a  triangle  varies  its  magnitude  only,  the  tensors  or 
ratios  of  the  sides  taken  two  and  two  remain  constant.  Hence 
the  tensors  or  ratios  of  the  sides  of  a  triangle  taken  two  and 
two  determine  the  form  of  the  triangle  but  not  its  magni- 
tude ;  i.e.^  they  determine  the  angles  but  not  the  sides. 

{n%  3 ;  197° ;  198° ;  201°) 

A  triangle,  which,  while  varying  its  size,  retains  its  form,  is 
sometimes  said  to  remain  similar  to  itself,  because  the  tri- 
angles due  to  any  two  stages  in  its  variation  are  similar  to 
one  another. 

213°.  In  the  right-angled  triangle  the  ratios  or  tensors  of 
the  sides  taken  in  pairs  are  important  functions  of  the  angles 

and  receive  distinctive  names. 
^'  ''       The  AOPM  is  right-angled  at  M, 

and  the  :lPOM  is  denoted  by  e. 

PM 
Then,    -~~  is  the  sine  of  ^,  and  is 

contracted  to  sin  B  in  writing. 

^^  is  the  sine  of  the  ^OPM,  but  as  ^OPM  is  the  comple- 
ment of  ^,  this  tensor  is  called  the  cosine  of  ^,  and  is  written  cos  Q. 

'-'  is  the  tangent  of  B  contracted  to  tan  B. 
OM  * 

Cor.  I.    V  ™  =  PM..OP,    .,  tan<»  =  liil^. 

OM     OP    OM'  cos» 


PM^.oM==op^ .-.  e)%(^y 


Cor.  2. 
or  sin-^-f  cos'-^^=i. 

214°.  LetOP'  =  OP  be  drawn  so  that  aP'OM'  =  POM  =  ^, 
and  let  P'M'  be  _L  on  OM. 

Then  z.P'OM  is  the  supplement  of  ^,  and  the  triangles 
P'OM'  and  POM  are  congruent. 


FUNCTIONS  OF  ANGLES. — AREAL  RELATIONS.     1 73 

sinP'OM  =  ^  =  ™^sin.. 

i.e.,  an  angle  and  its  supplement  have  the  same  sine. 

2.  CosP'OM  =  5^.     But  in  changing  from  OM  to  OM', 

on  the  same  line,  OM  vanishes  and  then  reappears  upon  the 
opposite  side  of  O. 

Therefore  OM  and  OM'  have  opposite  senses  (156°),  and 
if  we  consider  OM  positive,  OM'  is  negative. 

0M'=  -  OM,  and  hence 
an  angle  and  its  supplement  have  cosines  which  are  equal  in 
numerical  value  but  opposite  in  sign. 

215°,  Theorem. — The  area  of  a  parallelogram  is  the  pro- 
duct of  two  adjacent  sides  multiplied  by  the  sine  of  their 
included  angle.  (152°,!) 

AC  is  a  I  7  and  BP  is  X  upon  AD. 
Then  BP  is  the  altitude,  and  the 
area  =  AD.BP.  (153°,  i) 

But  BP  =  ABsin^BAP. 

£=7  =  AB .  AD  sin  ^BAP=rt<5  sin  ^. 

Cor.  I.  Since  the  area  of  a  triangle  is  one-half  that  of  the 
parallelogram  on  the  same  base  and  altitude, 

.'.  the  area  of  a  triangle  is  one-half  the  product  of  any  two 
sides  multiplied  by  the  sine  of  the  included  angle.     Or 
2A  =  ^<^sin  C  =  ^i:sin  A=<f^sin  B. 

216°.  Theorem. — The  area  of  any  quadrangle  is  one-half 
the  product  of  the  diagonals  multiplied  by  the  sine  of  the 
angle  between  them. 

ABCD  is  a  quadrangle  of  which  AC 
and  BD  are  diagonals. 

Let^AOB  =  ^  =  ^COD. 
Then  ^BOC=z.AOD  =  supp.  of  ^. 
AAOB  =^0A .  OB  sin  B, 
ABOC=iOB  .  CO  sin  B,     ACOD  =^0C .  DO  sin  B, 


74 


SYNTHETIC   GEOMETRY. 


ADOA=  |DO  .  OA  sin  6,  and  adding, 
Od.=:|AC.BDsin^. 


(compare  162°) 


217°.  BD  being  the  altitude  to  AC  in  the  AABC,  we  have 
B        from  172°,  2, 

But 

r2. 


AD  =  AB  cos  A=r  cos  A, 

a^  =  b^+C'-2bccos\. 
When  B  comes  to  B'  the  lK  becomes 

(214°,  2) 


D'     A  DC 

obtuse,  and  cos  A  changes  sign. 

If  we  consider  the  cosine  with  respect  to  its  magnitude 
only,  we  must  write  +  before  the  term  2bccosK,  when  A  be- 
comes obtuse.  But,  if  we  leave  the  sign  of  the  function  to  be 
accounted  for  by  the  character  of  the  angle,  the  form  given 
is  universal. 

Cor.  I.  ABCD  is  a  parallelogram.     Consider  the  AABD, 
then  '^T)'^  =  a'^-\-b^-2abcosd. 

Next,  consider  the  AABC.     Since 
^ABC  is  the  supplement  of  ^,  and 
BC  =  AD  =  /^, 
AO=d^  +  b'^  +  2abcosd. 
and  writing  these  as  one  expression, 
a'^->rb'^±2abcosd^ 
gives  both  the  diagonals  of  any  1 — 7,  one  of  whose  angles  is  0. 

Cor.  2.     DE  =  rt:cos^(CEbeingJ_to  AD),  CE  =  rtsin^. 
AE  =  ^  +  ^cos^; 
and     tanCAE  =  CE^_^zsin^ 
AE 


b  +  aco^  d' 
which  gives  the  direction  of  the  diagonal. 


218".  Def. — The  ratio  of  any  area  X  to  another  area  Y  is 
the  measure  of  X  when  Y  is  taken  as  the  unit-area,  and  is 

accordingly  expressed  as  — .     (Compare  i88°.) 

I.  Let  X  and  Y  be  two  similar  rectangles.     Then  X  =  ab 


FUNCTIONS  OF  ANGLES.— AREAL  RELATIONS.      175 

and  Y  =  a'd\  where  a  and  /;  are  adjacent  sides  of  the  nnX  and 
a'  and  d'  those  of  the  cziY. 

X  _a    d 

But  because  the  rectangles  are  similar,  —  =    . 

X^^  ^'     ^' 

Y     a"'' 
/.<?.,  the  areas  of  similar  rectangles  are  proportional  to  the 
areas  of  the  squares  upon  homologous  sides. 

2.  Let  X  and  Y  be  two  similar  triangles.     Then 

X  =  ^ad  sin  C,  Y = ha' d' sin  C, 

Y     c^b'     rt'^' 
because  the  triangles  are  similar,  (i97°) 

i.e.^  the  areas  of  similar  triangles  are  proportional  to  the 
areas  of  the  squares  upon  homologous  sides. 

3.  Let  X  denote  the  area  of  the  pentagon  ABCDE,  and  Y 
that  of  the  similar  pentagon  p^ 
A'B'CD'E'.     Then  A^\  R 

P^AD^      R^AC^ 

P'     A'D'2'    R'     A'C'^'  /  p    \     Q 

Q'     D'C'2*  ^ 

But  (X) 

P'     R'     Q'     P'  +  Q'  +  R' 
,  AD  _  AC  _  DC 

ana  aTD'-A'C'D'C" 

X_  DC^ 

Y     D'C'^' 
And  the  same  relation  may  be  proved  for  any  two  similar 
rectilinear  figures  whatever. 

.'.  the  areas  of  any  two  similar  rectilinear  figures  are  pro- 
portional to  the  areas  of  squares  upon  any  two  homologous 
lines. 


176  SYNTHETIC   GEOMETRY. 

4.  Since  two  circles  are  always  similar,  and  are  the  limits 
of  two  similar  regular  polygons, 

.*.  the  areas  of  any  two  circles  are  proportional  to  the  areas 
of  squares  on  any  homologous  chords  of  the  circles,  or  on 
line-segments  equal  to  any  two  similar  arcs. 

5.  When  a  figure  varies  its  magnitude  and  retains  its  form, 
any  similar  figure  may  be  considered  as  one  stage  in  its 
variation. 

Hence  the  above  relations,  i,  2,  3,  4,  may  be  stated  as 
follows  : — 

The  area  of  any  figure  with  coiista7it  form  varies  as  the 
square  upon  any  one  of  its  line-segments. 


Exercises. 

Two  triangles  having  one  angle  in  each  equal  have 
their  areas  proportional  to  the  rectangles  on  the  sides 
containing  the  equal  angles. 

Two  equal  triangles,  which  have  an  angle  in  each  equal, 
have  the  sides  about  this  angle  reciprocally  propor- 
tional, i.e.^  a  :  a'  =  b' :  b. 

The  circle  described  on  the  hypothenuse  of  a  right-angled 
triangle  is  equal  to  the  sum  of  the  circles  described  on 
the  sides  as  diameters. 

If  semicircles  be  described  outwards  upon  the  sides  of  a 
right-angled  triangle  and  a  semicircle  be  described  in- 
wards on  the  hypothenuse,  two  crescents  are  formed 
whose  sum  is  the  area  of  the  triangle. 

AB  is  bisected  in  C,  D  is  any  point  in      /::m^ 
AB,  and  the  curves  are  semicircles.    ^    q 
Prove  that  P  4-5  =  0 -f-R.  A  c    d 

If  ^,  b  denote  adjacent  sides  of  a  parallelogram  and  also 
of  a  rectangle,  the  ratio  of  the  area  of  the  parallelogram 
to  that  of  the  rectangle  is  the  sine  of  the  angle  of  the 
parallelogram. 


FUNCTIONS  OF  ANGLES! — AREAL  RELATIONS.   1 7/ 

7.  The  sides  of  a  concyclic  quadrangle  are  a,  b^  c^  d.     Then 

the  cosine  of  the  angle  between  a  and  b  is 
{d'^b^-c'-d'')l2{^ab^cd). 

8.  In  the  quadrangle  of  7,  \is  denotes  one-half  the  perimeter, 

area  =  ^{{s-d){s-  b)(s  -  c){s  ~d)}. 

9.  In  any  parallelogram  the  ratio  of  the  rectangle  on  the 

sum  and  differences  of  adjacent  sides  to  the  rectangle 
on  the  diagonals  is  the  cosine  of  the  angle  between  the 
diagonals. 

10.  U  a,  b  be  the  adjacent  sides  of  a  parallelogram  and  6  the 

angle  between  them,  one  diagonal  is  double  the  other 

when  cos^=-l-(^  +  ^\ 

11.  If  one   diagonal   of  a  parallelogram    is    expressed    by 

/  J  -^^^'—^ — H,  the  other  diagonal  is  n  times  as  long. 

12.  Construct  an  isosceles  triangle  in  which  the  altitude  is  a 

mean  proportional  between  the  side  and  the  base. 

13.  Three   circles    touch    two   lines   and   the  middle   circle 

touches  each  of  the  others.  Prove  that  the  radius  of 
the  middle  circle  is  a  mean  proportional  between  the 
radii  of  the  others. 

14.  In  an  equilateral  triangle  describe   three   circles  which 

shall  touch  one  another  and  each  of  w  hich  shall  touch 
a  side  of  the  triangle. 

15.  In  an  equilateral  triangle  a  circle  is  described  to  touch 

the  incircle  and  two  sides  of  the  triangle.  Show  that 
its  radius  is  one-third  that  of  the  incircle. 


PART   IV. 

SECTION  I. 

GEOMETRIC  EXTENSIONS. 

220°.  Let  two  lines  L  and  M  passing  through  the  fixed 
points  A  and  B  meet  at  P. 

When  P  moves  in  the  direction  of  the  arrow,  L  and  M 
approach  towards  parallelism,  and  the 
angle  APB   diminishes.      Since  the 
^         •-     .    *"  ~--^   lines  are  unlimited  (21°,  3)  P  may  re- 

cede from  A  along  L  until  the  segment  AP  becomes  greater 
than  any  conceivable  length,  and  the  angle  APB  becomes 
less  than  any  conceivable  angle. 

And  as  this  process  may  be  supposed  to  go  on  endlessly,  P 
is  said  to  "go  to  infinity^'  or  to  "be  at  infinity,"  and  the 
z_APB  is  said  to  vanish. 

But  lines  which  make  no  angle  with  one  another  are  parallel, 
.'.  Parallel  lines  meet  at  infinity^  and  lines  which  meet  at 
injitiity  are  parallel. 

The  symbol  for  "  infinity  "  is  00 . 

The  phrases  "  to  go  to  infinity,"  "  to  be  at  infinity,"  must 
not  be  misunderstood.  Infinity  is  not  a  place  but  a  property. 
Lines  which  meet  at  00  are  lines  so  situated  that,  having  the 
same  direction  they  cannot  meet  at  any  finite  point,  and 
therefore  cannot  meet  at  all,  within  our  apprehension,  since 
every  point  that  can  be  conceived  of  is  finite. 

178 


GEOMETRIC   EXTENSIONS. 


1/9 


The  convenience  of  the  expressions  will  appear  throughout 
the  sequel. 

Cor.  Any  two  lines  in  the  same  plane  meet :  at  a  finite 
point  if  the  lines  are  not  parallel,  at  infinity  if  the  lines  are 
parallel. 


221°.  L  and  M  are  lines  intersecting  in  O,  and  P  is  any 
point  from  which  PB  and  PA  are  || 
respectively  to  L  and  M.     A  third 
and  variable  line  N  turns  about  P 
in  the  direction  of  the  arrow. 

1.  AX.  BY  =  a  constant        (184') 

=  U  say. 

When  N  comes  to  parallelism 
with  L,  AX  becomes  infinite  and 
BY  becomes  zero. 

.-.  00.0  is  indefinite  since  U  may  have  any  value  we 
please. 

2.  The  motion  continuing,  let  N  come  into  the  position  N'. 
Then  AX'  is  opposite  in  sense  to  AX,  and  BY'  to  BY.  But 
AX  increased  to  00 ,  changed  sign  and  then  decreased  ab- 
solutely, until  it  reached  its  present  value  AX',  while  BY 
decreased  to  zero  and  then  changed  sign. 

/.  a  magnitude  changes  sign  when  it  passes  through  zero 
or  infinity. 

3.  It  is  readily  seen  that,  as  the  rotation  continues,  BY'  in- 
creases negatively  and  AX'  decreases,  as  represented  in  one 
of  the  stages  of  change  at  X"  and  Y".  After  this  Y"  goes  off 
to  00  as  X"  comes  to  A.  Both  magnitudes  then  change  sign 
again,  this  time  BY"  by  passing  through  00  and  AX'''  by  pass- 
ing through  zero. 

.Since  both  segments  change  sign  together  the  product  or 
rectangle  remains  always  positive  and  always  equal  to  the 
constant  area  U. 


l8o  SYNTHETIC    GEOMETRY. 

222°.  A  line  in  the  plane  admits  of  one  kind  of  varia- 
tion, rotation.  When  it  rotates  about  a  fixed  finite  point  it 
describes  angles  about  that  point.  But  since  all  the  lines  of 
a  system  of  parallels  meet  at  the  same  point  at  infinity,  rota- 
tion about  that  point  is  equivalent  to  translation,  without 
rotation,  in  a  direction  orthogonal  to  that  of  the  line. 

Hence  any  line  can  be  brought  into  coincidence  with  any 
other  line  in  its  plane  by  rotation  about  the  point  of  intersection. 

223".  If  a  line  rotates  about  a  finite  point  while  the  point 
simultaneously  moves  along  the  line,  the  point  traces  a  curve 
to  which  the  line  is  at  all  times  a  tangent.  The  line  is  then 
said  to  envelope  the  curve,  and  the  curve  is  called  the  en- 
velope of  the  line. 

The  algebraic  equation  which  gives  the  relation  between 
the  rate  of  rotation  of  the  line  about  the  point  and  the  rate  of 
translation  of  the  point  along  the  line  is  the  intrinsic  equation 
to  the  curve. 

224°.  A  line-segment  in  the  plane  admits  of  two  kinds  of 
variation,  viz.,  variation  in  length,  and  rotation. 

If  one  end-point  be  fixed  the  other  describes  some  locus 
depending  for  its  character  upon  the  nature  of  the  variations. 

The  algebraic  equation  which  gives  the  relation  between 
the  rate  of  rotation  and  the  rate  of  increase  in  length  of  the 
segment,  or  radius  vector,  is  the  polar  equation  of  the  locus. 

When  the  segment  is  invariable  in  length  the  locus  is  a 
circle. 

225°.  A  line  which,  by  rotation,  describes  an  angle  may 
rotate  in  the  direction  of  the  hands  of  a  clock  or  in  the  con- 
trary direction. 

If  we  call  an  angle  described  by  one  rotation  positive  we 
must  call  that  described  by  the  other  negative.  Unless  con- 
venience requires  otherwise,  the  direction  of  rotation  of  the 
hands  of  a  clock  is  taken  as  negative. 


GEOMETRIC   EXTENSIONS. 


I8l 


An  angle  is  thus  counted  from  zero  to  a  circumangle  either 
positively  or  negatively. 

The  angle  between  AB  and  A'B'  is  the 
rotation  which  brings  AB  to  A'B',  and 
is  either  +  a  or  —  j3,  and  the  sum  of  these  A 
two  angles  irrespective  of  sign  is  a  cir- 
cumangle. 

When  an  angle  exceeds  a  circumangle  the  excess  is  taken 
in  Geometry  as  the  angle. 

Ex.  QA  and  QB  bisect  the  angles  CAB  and  ABP  extern- 
ally ;  to  prove  that  z_P  =  2^Q. 

The  rotation  which  brings  CP 
to  AB  is  -20,  AB  to  BP  is  +2^9, 
.-.   ^P  =  2(/3-a). 

Also,  the  rotation  which  brings 
AQ  to  AB  is  -a,  and  AB  to  BO 
is  +/3,  .-.  ^Q=/3-a. 

z.(CP.BP)  =  2Z-(AQ.BQ). 

This  property  is  employed  in  the  working  of  the  sextant. 


226°.  Let  AB  and  CD  be  two  diameters  at  right  angles. 
The  rectangular  sections  of  the  plane 
taken  in  order  of  positive  rotation  and 
starting  from  A  are  called  respectively 
the  first,  second,  third,  and  fourth 
quadrants,  the  first  being  AOC,  the  ^\ 
second  COB,  etc. 

The  radius  vector  starting  from  co- 
incidence with  OA  may  describe  the 
positive  zJ\OP,  or  the  negative  Z.AOP'. 
Let   these   angles  be  equal   in  absolute  value,  so  that  the 
AMOPs  AMOP',  PM  being  ±  on  OA. 

Then   PM=-P'M,  since  in  passing  from  P  to   P',  PM 
passes  through  zero. 


sinAOP'  = 


P'M 
OP 


-PM 
OP 


=  -sinAOP. 


82 


SYNTHETIC   GEOMETRY. 


and 


cosAOP'=OM^=0^=cosAOP. 


.'.  the  sine  of  an  angle  changes  sign  when  the  angle  does, 
but  the  cosine  does  not. 

227°.  As  the  angle  AOP  increases,  OP  passes  through  the 
several  quadrants  in  succession. 

When  OP  lies  in  the  ist  Q.,  sin  AOP  and  cos  AOP  are 
both  positive  ;  when  OP  lies  in  the  2nd  g.,  sin  AOP  is  posi- 
tive and  cos  AOP  is  negative ;  when  OP  lies  in  the  3rd  O., 
the  sine  and  cosine  are  both  negative  ;  and,  lastly,  when  OP 
lies  in  the  4th  ^.,  sin  AOP  is  negative  and  the  cosine 
positive. 

Again,  when  P  is  at  A,  i.AOP  =  o,  and  PM==o,  while 
OM  =  OP.  .•.  sin  0  =  0  and  cos  0=1. 

When  P  comes  to  C,  PM  =  OP  and  0M=o,  and  denoting 
a  right  angle  by  \  (207°,  Cor.  4) 

sin  2=  I,  and  cos^  =  o. 

When  P  comes  to  B,  PM=o  and  OM=  - 1, 
sin  7r  =  o,  and  cos7r=  —  I. 

Finally  when  P  comes  to  D,  PM=  -OP  and  OM=o. 

sin  ^'^=  -  I,  and  cos  ^^^  =  0. 

•7.  '  -? 


These  variations  of  the  sine  and  cosine  for  the  several 
quadrants  are  collected  in  the  following  table  : — 


ISt(?. 

2nde. 

3rd<>. 

4th  (>. 

Sine, 

Cosine,     .... 

+ 

+ 

+ 

- 

+ 

Sine, 

Cosine,      .... 

From      To 
I          0 

From      To 
I          0 
0    -T 

From       To 

0  -  r 
- 1       0 

From       To 

-  I          0 

0          I 

GEOMETRIC   EXTENSIONS.  183 

228°.  ABC  is  a  triangle  in  its  circumcircle  whose  diameter 
we  will  denote  by  d.  ^ ^^        ^b 

Let  CD  be  a  diameter. 
Then  ^D  =  Z.A,  (106°,  Cor.  i) 

and    Z.CBD=~|-  ^^ 

CB  =  CDsinCDB  =  ^sinA  =  ^. 
and  from  symmetry, 

<!/=    ^     _    b     ^    c 
sin  A     sin  B     sin  C* 

Hence  the  sides  of  a  triangle  are  proportional  to  the  sines  of 
the  opposite  angles  ;  and  the  diameter  of  the  circumcircle  is 
the  quotient  arising  from  dividing  any  side  by  the  sine  of  the 
angle  opposite  that  side. 


PRINCIPLE  OF  ORTHOGONAL  PROJECTION. 

229°.  The  orthogonal  projection  {i6y°,  2)  of  PO  on  L  is 
P'O',  the  segment  intercepted  between 
the  feet  of  the  perpendiculars  PP'  and 
QQ'. 

Now      P'Q'  =  PQcos(PQ  .  P'Q'). 
.'.  the  projection  of  any  segment  on  a 
given  line  is  the  segment  multiplied  by  the  cosine  of  the 
angle  which  it  makes  with  the  given  line. 

From  left  to  right  being  considered  as  the  +  direction 
along  L,  the  segment  PQ  lies  in  the  ist  Q.,  as  may  readily  be 
seen  by  considering  P,  the  point  from  which  we  read  the 
segment,  as  being  the  centre  of  a  circle  through  Q. 

Similarly  QP  lies  in  the  3rd  Q.,  and  hence  the  projection 
of  PO  on  L  is  +  while  that  of  OP  is  -. 

When  PQ  is  J_  to  L,  its  projection  on  L  is  zero,  and  wheTi 
II  to  L  this  projection  is  PO  itself. 

Results  obtained  through  orthogonal  projection  are  univer- 
sally true  for  all  angles,   but   the  greatest   care   must   be 


1 84 


SYNTHETIC   GEOMETRY. 


exercised   with   regard  to   the   signs   of   angular    functions 
concerned. 

Ex.  AX  and  OY  are  fixed  lines  at  right  angles,  and  AQ  is 
any  line  and  P  any  point. 

Required  to  find  the  ±PQ  in  terms  of 
AX,  PX,  and  the  z_A. 

Take  PO  as  the  positive  direction,  and 
project  the  closed  figure  POAXP  on  the 
line  of  PO.     Then 
pr.PQ  +  pr.QA  +  pr.AX  +  pr.XP  =  o.  (i68') 
Now,  pr.PQ  is  PQ,  and  pr.QA  =  o ;  AX  lies  in  ist  Q.^  and 
XP  in  the  3rd  Q. 

Moreover  z.(AX .  PQ),  i.e.,  the  rotation  which  brings  AX  to 
PQ  in  direction  is  -Z-N,  and  its  cosine  is  +. 
cos4AX.PQ)=+sinA. 
Also,  pr.XP  is  -XPcos^XPQ=  -XPcosA. 
PO  =  XPcosA-AXsinA. 


SIGNS  OF  THE  SEGMENTS  OF  DIVIDED 
LINES  AND  ANGLES. 


230°.  AOB  is  a  given  angle  and  ^AOB=  -Z.BOA. 
Let  OP  divide  the  Z.AOB  internally,  and  OQ  divide  it  ex- 
ternally into  parts  denoted  respectively 
by  a,  /3,  and  a!^. 

If  a  is  the  ^AOP  and  /3  the  ^POB,  a 
and  /3  are  both  positive.     But  if  we 
write    Q    for     P,    a!  =  /-AOQ,     and 
/3'  =  ^Q0B,  and  a  and  /S'  have  contrary  signs. 

On  the  other  hand,  if  a  is  ^AOP  and  ^  the  ^BOP,  a  and  /8 
have  contrary  signs,  while  replacing  P  by  Q  gives  a  and  ^ 
with  hke  signs. 

The  choice  between  these  usages  must  depend  upon  con- 
venience :  and  as  it  is  more  symmetrical  with  a  two-letter 


GEOMETRIC    EXTENSIONS.  1 85 

notation  to  write  AGP,  BOP,  AGO,  BOO,  than  AGP,  FOB, 
etc.,  we  adopt  the  convention  that  internal  division  of  an 
angle  gives  segments  with  opposite  signs,  while  external 
division  gives  segments  with  like  signs. 

In  like  manner  the  internal  division  of  the  segment  AB 
gives  parts  AP,  BP  having  unlike  signs,  while  external  divi- 
sion gives  parts  AQ,  BQ  having  hke  signs. 

Def. — A  set  of  points  on  a  line  is  called  a  7-ajige,  and  the 
line  is  called  its  axis. 

By  connecting  the  points  of  the  range  with  any  point  not 
on  its  axis  we  obtain  a  corresponding  pencil.  (203°,  Def) 

Cor.  To  any  range  corresponds  a  pencil  for  every  vertex, 
and  to  any  vertex  corresponds  a  range  for  every  axis,  the 
axis  being  a  transversal  to  the  rays  of  the  pencil. 

If  the  vertex  is  on  the  axis  the  rays  are  coincident ;  and  if 
the  axis  passes  through  the  vertex  the  points  are  coincident. 

231°.  BY  is  any  line  dividing  the  angle  B,  and  CR,  AP  are 
perpendiculars  upon  BY.  B 

Then         AAPY^ACRY, 
and  AP  is  AB  sin  ABY, 

and  CR  is  BC  sin  CBY, 

AY^AB  sin  ABY 
CY  CB  •  sin  CBY' 
Therefore  a  line  through  the  vertex  of  a  tri- 
angle' divides  the  base  into  segments  which 
are  proportional  to  the  products  of  each  conterminous  side 
multiplied  by  the  sine  of  the  corresponding  segment  of  the 
vertical  angle. 

Cor.  I.  Let  BY  bisect  /.B,  then  ^^=-. 

YC     a 

AY=^(^-AY),  and  AY  =  — . 

Thence  YC  =  -^. 

a  +  c 


1 86  SYNTHETIC   GEOMETRY. 

Which  are  the  segments  into  which  the  bisector  of  the  ^B 
divides  the  base  AC. 

Cor.  2.  In  the  AABY, 

BY2  =  ABHAY--2AB.AY.cosA.  (217°) 

But  cos  A  =  ^'±^'-":-''-  (217^),  and  AY=-^-, 

ibc  "  a  +  c 

whence  by  reduction 

which  is  the  square  of  the  length  of  the  bisector. 

Cor.  3.  When  AY  =  CY,  BY  is  a  median,  and 

AB^sinYBC 

CB     sinABY" 
.'.a  median  to  a  triangle  divides  the  angle  through  which  it 
passes  into  parts  whose  sines  are  reciprocally  as  the  con- 
terminous sides. 

232°.  In  any  range,  when  we  consider  both  sign  and  mag- 
nitude, the  sum    AB  +  BC  +  CD  +  DE  +  EA=o, 
however  the  points  may  be  arranged. 

For,  since  we  start  from  A  and  return  to  A,  the  translation 
in  a  +  direction  must  be  equal  to  that  in  a  -  direction. 

That  this  holds  for  any  number  of  points  is  readily  seen. 

Also,  in  any  pencil,  when  we  consider  both  sign  and  mag- 
nitude, the  sum  zA0B  +  ^B0C-hz.COD-f^D0A=o. 

For  we  start  from  the  ray  OA  and  end  with  the  ray  OA, 
and  hence  the  rotation  in  a  -f-  direction  is  equal  to  that  in  a 
-  direction. 


RANGES  AND  PENCILS  OF  FOUR. 

533°.  Let  A,  B,  C,  P  be  a  range  of  four,  then 
AB.CP-l-BC.AP  +  CA.BP  =  o. 


GEOMETRIC   EXTENSIONS.  1 8/ 

Proof.—  AP  =  AC  +  CP,  and  BP  =  BC  +  CP. 

.'.  the  expression  becomes 

BC(AC  +  CA) +  (AB  +  BC  +  CA)CP, 
and  each  of  the  brackets  is  zero  (232°).     .'.  etc. 


234°.  Let  O  .  ABCP  be  a  pencil  of  four.     Then 
sinAOB.sinCOP  +  sinBOC.sinAOP  +  sinCOA.sinBOP^o. 

P^^^y:_/^AOB=|OA .  OB  sin  AOB, 
also  AAOB=^AB.;^, 

where  p  is  the  common  altitude  to  all 
the  triangles. 

AB  ./  =  OA .  OB .  sin  AOB. 
Similarly,     CP./=OC.  OP.  sin  COP.     '^      ^         c\ 

AB .  CP  .^2==  OA .  OB  .  OC .  OP  sin  AOB  .  sin  COP. 
Now,  /"  and  OA .  OB .  OC .  OP  appear  in  every  homologous 
product,  .•.(AB.CP  +  BC.AP  +  CA.BP)/^ 

=  OA .  OB .  OC .  OP(sin  AOB .  sin  COP 

+  sin  BOC .  sin  AOP  +  sin  COA .  sin  BOP). 
But  the  bracket  on  the  left  is  zero  (233°),  and  OA.  OB.OC.OD 
is  not  zero,  therefore  the  bracket  on  the  right  is  zero.       q.e.d. 


235''.  From  P  let  perpendiculars  PA',  PB',  PC  be  drawn  to 
OA,  OB,  and  OC  respectively.     Then 

sinAOP  =  ^,  sinBOP=^^,  etc., 

and  putting  these  values  for  sin  AOP,  etc.,  in  the  relation  of 
234°,  we  have,  after  multiplying  through  by  OP, 

C'P.sinAOB  +  A'P.sinBOC  +  B'P.sinCOA  =  o. 
Or,  let  L,  M,  and  N  be  any  three  concurrent  lines,  /,  ///,  7i 
the  perpendiculars  from  any  point  P  upon  L,   M,  and   N 
respectively,  then 

>^       /sin  MN+;;zsinNL  +  ;;sin  LM=o. 
where  MN  denotes  the  angle  between  M  and  N,  etc. 


SYNTHETIC   GEOMETRY. 

Let  four  rays  be  disposed  in  the  order  OA, 
-P  OB,  OC,  OP,  and  let  OP  be  perpendicular 
to  OA. 

Denote  /.AOC  by  A,  and  ^AOB  by  B. 
Then  234°  becomes 

sinBcosA  +  sin(A-B)sin2-sinAcosB=o, 
or,      sin(A-B)=sinAcosB-cosAsinB. 
Similarly,  by  writing  the  rays  in  the  same  order  and  making 
^BOP  a  n,  and  denoting  ^AOB  by  A  and  ^BOC  by  B,  we 
obtain  sin  (A  +  B) = sin  A  cos  B  +  cos  A  sin  B. 

Also,  by  writing  the  rays  in  the  order  OA,  OP,  OB,  OC, 
and  denoting  z_AOP  by  A  and  Z.BOC  by  B,  we  obtain 

(i)  when  ^AOB=n, 

cos  (A  -  B)= cos  A  cos  B  +  sin  A  sin  B  ; 

(2)  when  zAOC=n, 

cos  (A  +  B)  =  cosAcosB-sinAsinB  ; 
which  are  the  addition  theorems  for  the  sine  and  cosine. 

Ex.  2.  ABC  is  a  triangle  and  P  is  any  point.     Let  PX, 
p  PY,    PZ  be  perpendiculars    upon    BC, 

K-^.  ^ A    CA,  AB,  and  be  denoted  by  P«,  P^,  Pc 


1       \  ''""-v//-    I'espectively. 

I  /^/     I    Then  (235°) 

X        B       c      D         PQsinA  +  PYsinB  +  PZsinC  =  o. 
But  if  AD  is±to  BC,  AD  =  ^sin  C  =  QX. 

(PX-(^sinC)sinA  +  PYsinB  +PZsinC  =  o, 
or  .2:(P„sinA)  =  <^sinAsinC. 

Similarly,  S(PaSin  A)  =  rsin  Bsin  A 

=  ^sin  C  sin  B, 
2(P„sin  A)=^{abcs\n^K  sin2B  sin^C}. 

Hence  the  function  of  the  perpendicular 

P,,  sin  A  +  Pj  sin  B  +  Pc  sin  C 
is  constant  for  all  positions  of  P.     This   constancy  is   an 
important  element  in  the  theory  of  trilinear  co-ordinates. 


GEOMETRIC   EXTENSIONS. 


189 


237".  A,  B,  C  being  a  range  of  three,  and  P  any  point  not 
on  the  axis, 
AB.CP-4-BC.AP-  +  CA.BP- 

--AB.BC.CA. 

Proof.— LtX  P()  be  X  to  AC.     Then 
AO  =  AC  +  CQ,   BQ  =  BC  +  CQ, 
and  the  expression  becomes 

(AB  +  BC  +  CA)(PQ2  +  CQ-)  +  BC.CA(BC-AC) 

=  BC .  CA(BC  +  CA)  =  BC .  CA .  BA 
=  -AB.BC.CA. 


Exercises. 

1.  A  number  of  stretched  threads  have  their  lower  ends  fixed 

to  points  lying  in  line  on  a  table,  and  their  other  ends 
brought  together  at  a  point  above  the  table.  What  is 
the  character  of  the  system  of  shadows  on  the  table 
when  {a)  a  point  of  light  is  placed  at  the  same  height 
above  the  table  as  the  point  of  concurrence  of  the 
threads  ?  {b)  when  placed  at  a  greater  or  less  height  t 

2 .  If  a  line  rotates  uniformly  about  a  point  while  the  point 

moves  uniformly  along  the  line,  the  point  traces  and 
the  line  envelopes  a  circle. 

3.  If  a  radius  vector  rotates  uniformly  and  at  the  same  time 

lengthens  uniformly,  obtain  an  idea  of  the  curve  traced 
by  the  distal  end-point. 

4.  Divide  an  angle  into  two  parts  whose  sines  shall  be  in  a 

given  ratio.     (Use  231°,  Cor.  3.) 

5.  From  a  given  angle  cut  off  a  part  whose  sine  shall  be  to 

that  of  the  whole  angle  in  a  given  ratio. 

6.  Divide  a  given  angle  into  two  parts  such  that  the  product 

of  their  sines  may  be  a  given  quantity.  Under  what 
condition  is  the  solution  impossible  ? 

7.  Write  the  following  in  their  simplest  form  : — 

sin(7r-^),  sin  (2+^),  sin  {tt- (^-|) }, 

C0s(27r  +  ^),  cos  -j27r-(^-|)},  COS  {  ^  -  (J-)-^)  }. 


IQO 


SYNTHETIC   GEOMETRY. 


8.  Make  a  table  of  the  variation  of  the  tangent  of  an  angle 

in  magnitude  and  sign. 

9.  OM  and  ON  are  two  lines  making  theiLMON  =  w,  and 

PM  and  PN  are  perpendiculars  upon  OM  and  ON 

respectively.     Then  OP  sin  a;  =  MN. 
A  transversal  makes  angles  A',  B',  C  with  the  sides  BC, 

CA,  AB  of  a  triangle.     Then 

sin  A  sin  A'  +  sin  B  sin  B'  +  sin  C  sin  C = o. 
OA,  OB,  OC,  OP  being  four  rays  of  any  length  whatever, 
AAOB.ACOP+ABOC.AAOP+ACOA.ABOP=o. 
If  r  be  the  radius  of  the  incircle  of  a  triangle,  and  r-^  be 

that  of  the  excircle  to  side  a^  and  if  p^  be  the  altitude 

to  the  side  a^  etc., 

.^  .(sin A  +  sinB  +  sinC) 
sni  A 

-^-\  (-sin  A  +  sinB  +  sinC), 
sm  A 


10 


II. 


J  2, 


13- 


A 


and 


+  -+-  = 


^+^ 


^3    P\    Pi    /; 


+  ^  =  i 


(Use  235°.) 


The  base  AC  of  a  triangle  is  trisected  at  M  and  N,  then 
BN-  =  i(3BC-  +  6BA2-2AC). 


SECTION   II. 


CENTRE  OF  MEAN  POSITION. 


A,  B,  C,  D  are  any  points  in  line,  and  perpendiculars 
AA',  BB',  etc.,  are  drawn  to  any  fixed 
line  L.  Then  there  is,  on  the  line, 
evidently  some  point,  O,  for  which 
AA'  +  BB'  +  CC'+DD'  =  40N  ; 
and  ON  is  less  than  A  A'  and  greater 
than  DD'. 

The  point  O  is  called  the  centre  o/jnean  position,  or  simply 
the  7ne(m  centi'e,  of  the  system  of  points  A,  B,  C,  D. 


CENTRE   OF    MEAN    POSITION.  IQI 

Again,  if  we  take  multiples  of  the  perpendiculars,  as  a.AA', 
b  .  BB',  etc.,  there  is  some  point  O,  on  the  axis  of  the  points, 
for  which 

^?.AA'  +  /5.BB'  +  r.CC'  +  ^.  DI)'  =  (rt  +  /A+6-+^/)ON. 
Here  again  ON  lies  between  AA'  and  DD'. 

O  is  then  called  the  nieati  centre  of  the  system  of  points  for 
the  system  of  multiples. 

Def.—  For  a  range  of  points  with  a  system  of  multiples  we 
define  the  mean  centre  by  the  equation 

2(^?.A0)  =  o, 
where  2(/t .  AO)  is  a  contraction  for 

^.AO  +  ^.BO  +  6-.CO  +  ..., 
and   the   signs   and  magnitudes  of  the  segments  are  both 
considered. 

The  notion  of  the  mean  centre  or  centre  of  mean  position 
has  been  introduced  into  Geometry  from  Statics,  since  a 
system  of  material  points  having  their  weights  denoted  by  «, 
b,  c,  ...,  and  placed  at  A,  B,  C,  ...,  would  "balance"  about 
the  mean  centre  O,  if  free  to  rotate  about  O  under  the  action 
of  gravity. 

The  mean  centre  has  therefore  a  close  relation  to  the 
"  centre  of  gravity  "  or  "mass  centre"  of  Statics. 

239°.  Theorem. — If  P  is  an  independent  point  in  the  line  of 

any  range,  and  O  is  the  mean  centre,      ^       ^  

S(^ .  AP)  =  S(«) .  OP.  "a      b     S      c  p 

/•r^^/-- AP  =  AO  +  OP,  BP  =  BO  +  OP,  etc., 

rt.AP  =  ^.AO  +  rt.OP,    ^.BP  =  <^.BO  +  ^.OP,  etc. 

2(^ .  AP)  =  Z(^ .  AO)  +  2(^) .  OP. 
But,  if  O  is  the  mean  centre, 

2(<z.  AO)=o,  by  definition, 
2(rt.AP)=S(.?)-OP. 

Ex.  The  mean  centre  of  the  basal  vertices  of  a  triangle 


192  SYNTHETIC   GEOMETRY. 

when  the  multiples  are  proportional  to  the  opposite  sides  is 
the  foot  of  the  bisector  of  the  vertical  angle. 

240°.  Let  A,  B,  C,  ...  be  a  system  of  points  situated  any- 
where in  the  plane,  and  let  AL,  BL,  CL,  ...,  AM,  BM,  CM, 
...,  denote  perpendiculars  from  A,  B,  C,  ...  upon  two  Hnes  L 
and  M. 

Then  we  define  the  mean  centre  of  the  system  of  points  for 
a  system  of  multiples  as  the  point  of  intersection  of  L  and  M 
when  2(^z.AL)  =  o, 

and  2(rt:.AM)=o. 

If  N  be  any  other  line  through  this 
centre,  S(^.AN)=o. 

For,  let  A  be  one  of  the  points. 
Then,  since  L,  M,  N  is  a  pencil  of  three 
and  A  any  point,  (235°) 

AL .  sin  MON  +  AM  .  sin  NOL+AN  .  sin  LOM  =0, 
also        BL  .  sin  MON  +  BM  .  sin  NOL  +  BN .  sin  LOM  =  0, 

and  multiplying  the  first  b}'  <?,  the  second  by  b^  etc.,  and  adding, 
2(^.AL)sinMON  +  2(^.AM)sinNOL  +  2(rt.AN)sinLOM=o. 
But  2:(^.  AL)  =  S(rt'.  AM)  =  o,  by  definition, 

2:(rt.AN)  =  o. 

241".    'J'heore7n. — If  O  be  the  mean  centre  of  a  system  of 
points  for  a  system  of  multiples,  and  L  any  line  whatever, 
20?.AL)  =  2(rt).OL. 

Proof. — Let  M  be  ||  to  L  and  pass  through  O.     Then 
AL  =  AM  +  ML,  .-.  ^.AL=-^.AM  +  ^?.ML, 

BL  =  BM  +  ML,  ,-.   /;.BL  =  ^.BM  +  /5.ML, 


adding,  ^{a .  AL)  =  Y.{a .  AM)  +  ^{d) .  ML. 

But,  since  M  passes  through  O, 

S(«.AM)=:o,  and  ML  =  OL, 

S(^.AL)=S(^).OL. 


CENTRE   OF   MEAN    POSITION.  1 93 

242°.   Theorem. — The  mean   centre   of  the  vertices  of  a 


triangle  with  multiples  proportional  to 
the  opposite  sides  is  the  centre  of  the 
incircle. 

Proof. — Take  L  along  one  "of  the  sides, 
as  BC,  and  let/  be  the  ±  from  A.     Then  A 

2(^.AL)  =  ^./ 
and  S(.i:).OL  =  («  +  <^+<:).OL, 

■•■(^39°)  OL  =  -^^=f  =  .;      (.53',  Ex..) 

i.e.^  the  mean  centre  is  at  the  distance  r  from  each  side,  and 
is  the  centre  of  the  incircle. 

Cor.  I.  If  one  of  the  multiples,  as  a,  be  taken  negative, 

OL=-~^f     =z:A^  _^'         (j53°  Ex.  2) 

/>.,  the  mean  centre  is  beyond  L,  and  is  at  the  distance  r'  from 
each  side,  or  it  is  the  centre  of  the  excircle  to  the  side  a. 

Cor.  2.  If  any  line  be  drawn  through  the  centre  of  the  in- 
circle of  a  triangle,  and  a,  j8,  7  be  the  perpendiculars  from  the 
vertices  upon  it,  aa  +  d^  +  cy=o, 

and  if  the  line  passes  through  the  centre  of  an  excircle,  that 
on  the  side  ^  for  example,     aa  =  b^  +  cy. 


Exercises. 

If  a  line  so  moves  that  the  sum  of  fixed  multiples  of  the 
perpendiculars  upon  it  from  any  number  of  points  is 
constant,  the  line  envelopes  a  circle  whose  radius  is 

The  mean  centre  of  the  vertices  of  a  triangle,  for  equal 

multiples,  is  the  centroid. 
The  mean  centre  of  the  vertices  of  any  regular  polygon,  for 

equal  multiples,  is  the  centre  of  its  circumcircle. 

N 


194  SYNTHETIC   GEOMETRY. 

243°.  Theorem. — If  O  be  the  mean  centre  of  a  system  of 
points  for  a  system  of  multiples,  and  P  any  independent  point 
in  the  plane, 

Proof. — Let  O  be  the  mean  centre,  P  the 
independent  point,  and  A  any  point  of  the 
system.  Let  L  pass  through  O  and  be  per- 
pendicular to  OP,  and  let  AA'  be  perpen- 
dicular to  OP.  Then 
AP2=AOHOP2-20P.OA', 
and  «.AP2  =  ^.A02  +  «.OP2-20P.^.OA'. 

Similarly    <5.  BP2  =  <^.  30^  +  ^.  OP2-20P.^.  OB', 


2(^.AP2)  =  2(«.A02)-l-2(^).0P2-20P.2(^.0A'). 
But        2(^.0A')  =  2(«.AL)=o,  (241°) 

2(^.  AP2)  =  2(^.  A02)  +  2(rt).  0P2.  q.e.d. 

Cor.  In  any  regular  polygon  of  ;^  sides  -th  the  sum  of  the 

squares  on  the  joins  of  any  point  with  the  vertices  is  greater 
than  the  square  on  the  join  of  the  point  with  the  mean  centre 
of  the  polygon  by  the  square  on  the  circumradius. 
For  making  the  multiples  all  unity, 

2(AP2)  =  /z^  +  «OP2, 

i2(AP2)  =  OP2  +  r2. 

71 

Ex.  Let  a,  b,  c  be  the  sides  of  a  triangle,  and  a,  j8,  7  the 
joins  of  the  vertices  with  the  centroid.     Then  (242°,  Ex.  2) 
2(AP2)=2(A02)-f-30P2. 
ist.    Let  P  be  at  A,         ^^.^^^^a^  +  zS^-l-T'+Sa', 
2nd.     „    P      „     B,         r^  +  ^^^a^  +  iSHv'  +  S^', 
3rd.      „    P      „     C,         «2  +  <^^  =  a2  +  /32-f-7H3r, 
whence  Vz^+(^^  +  c^=3(a-+/3H7-). 

Ex.  If  ABCDEFGH  be  the  vertices  of  a  regular  octagon 
taken  in  order,  AC2  +  ADHAE2-t-AF2  +  AG2=i4r2. 


CENTRE  OF   MEAN   POSITION.  195 

244°.  Let  O  be  the  centre  of  the  incircle  of  the  AABC  and 
let  P  coincide  with  A,  B,  and  C  in  succession. 

ist.  bc"^  cl^  =  ^{a .  AO^)  +  2(rt)A02, 

2nd.      ac^  +t-«2_2(^.A02)  +  S(rt)B02, 

3rd.       aB'-^ba'^         =2(^.AO2)  +  2(.0CO2. 

Now,  multiply  the  ist  by  a,  the  2nd  by  b^  the  3rd  by  c,  and 

add,  and  we  obtain,  after  dividing  by  {a^rb^rc\ 

^{a.  AO^) =abc. 

Cor.  I,  For  any  triangle,  with  O  as  the  centre  of  the  incircle, 
the  relation       2(^.  AP2)=S(rt.  A02)+2(rt)OP2 
becomes  'Z{a.AV^)=abc+2s.  OP^, 

and,  if  O  be  the  centre  of  an  excircle  on  side  ^,  for  example, 

S(^.  AP2)=  -abc+2{s-a)0V\ 
where  a  denotes  that  a  alone  is  negative. 

Cor.  2.  Let  P  be  taken  at  the  circumcentre,  and  let  D  be 
the  distance  between  the  circumcentre  and  the  centre  of  the 
incircle.     Then  AP  =  BP  =  CP  =  R. 

2s'K^=^abc  +  2sT>\ 
But  abc=^/S^,  (204°,  Cor.) 

and  s=^,  (153°,  Ex.  i) 

D2=R2-2Rr. 

Cor.  3.  If  Dj  be  the  distance  between  the  circumcentre  and 
the  centre  of  an  excircle  to  the  side  a,  we  obtain  in  a  similar 
manner  D  ^  =  R^  +  2  Rr^. 

Similarly  D2^  =  RH2Rr2, 

D32  =  R2  +  2Rr3. 

245°.  Ex.  To  find  the  product  OA.  OB .  DC,  where  O  is 
the  centre  of  the  incircle. 

Let  P  coincide  with  A.     Then  (244°) 
bc''  +  ^c=abc+2s.K0\ 


196  SYNTHETIC   GEOMETRY. 

Similarly  B0^=£%1^\  and  C02  =  ^%l£), 

AO^.  BO^.  CO^: 


s  s 

_  a^^c^s(s  -  a)(s  -  /?){s  -  c) 


=^A'=i6RV, 
and  OA.OB.OC  =  4R^2. 

246°.  If  S(rt .  AP^)  becomes  constant,  /^,  we  have 

^  =  2(^.A02)  +  2(rt)0P^ 

and  2(^z .  AO^)  being  independent  of  the  position  of  P,  and 

therefore  constant  for  variations  of  P,  OP  is  also  constant, 

and  P  describes  a  circle  whose  radius  is 

p2_>^-S(^.AO^) 

.'.  If  a  point  so  moves  that  the  sum  of  the  squares  of  its 
joins  with  any  number  of  fixed  points,  each  multiplied  by  a 
given  quantity,  is  constant,  the  point  describes  a  circle 
whose  centre  is  the  mean  point  of  the  system  for  the  given 
multiples. 

Exercises, 

1.  If  O',  O",  O'"  be  the  centres  of  the  escribed  circles, 

AO'.BO".CO"'=4R«5-'. 

2.  A0'.B0'.C0'  =  4R^i'. 

3.  s.OL  =  {s-  a)0'L  +  {s-  ^)0"L  +  {s-  c)0"'L, 
where  L  is  any  line  whatever. 

4.  If  P  be  any  point, 

5.  (s-a).  O'O^  +  {s-d).  0"02  +  {s~  c)0"'0''  =  labc. 

6.  If  D  is  the  distance  between  the  circumcentre  and  the 

centroid,  Ti"  =\{c)P:- -  a^  -  V"  -  c"). 


OF  COLLIN.EARITY  AND  CONCURRENCE.       I9; 

SECTION  III. 

OF  COLLINEARITY  AND  CONCURRENCE. 

247°.  Def.  I. — Three  or  more  points  in  line  are  collinear, 
and  three  or  more  lines  meeting  in  a  point  are  concurrent. 

Def.  2. — A  tetragram  or  general  quadrangle  is  the  figure 
formed  by  four  lines  no  three  of 
which  are  concurrent,  and  no  two 
of  which  are  parallel. 

Thus  L,  M,  N,  K  form  a  tetra-  l/  ^^^^/c 

gram.     A,  B,  C,  D,  E,  F  are  its  six 
vertices.    AC,  BD  are  its  internal     ^^ 
diagonals,  and  EF  is  its  external  diagonal. 

248°.  The  following  are  promiscuous  examples  of  collinear- 
ity  and  concurrence. 

Ex.  I.  AC  is  a  1     7,  and  P  is  any  point.     Through  P,  GH 

is  drawn  |1  to  BC,  and  EF  ||        k N  o 

to  AB. 

The  diagonals  EG,  HF, 
and  DB  of  the  three  /  7s 
AP,  PC,  and  AC  are  concur- 
rent. 

EG  and  HF  meet  in  some    D  H  c         L 

point  O  ;  join  BO  and  complete  the  ZZZ70KDL,  and  make 
the  extensions  as  in  the  figure. 

We  are  to  prove  that  D,  B,  O  are  collinear. 

/'r^<?/:— /ZZ7KG  =  Z=7GM,  and  ^Z7FL  =  £i:7FN,   (145°) 
Ei:7KG  =  ^=Z7GF  +  ZZ:7BM, 
and  z=:7FL  =  £:Z7GF  +  z:=7NB. 

Hence  EI::7KB  =  CZ7BL, 

B  is  on  the  line  DO,  (i45°j  Cor.  2) 

and  D,  B,  O  are  collinear. 


198 


SYNTHETIC  GEOMETRY. 


Ex.  2.  The  middle  points  of  the  three  diagonals  of  a  tetra- 

gram  are  collinear. 

ABCDEF  is  the  tet- 
ragram,  P,  Q,  R  the 
middle  points  of  the 
diagonals. 

Complete  the  paral- 
lelogram AEDG,  and 
through  B  and  C  draw 
lines  II  to  AG  and  AE 
respectively,  and  let 
them  meet  in  T. 

Then,  from  Ex.  i, 
IH  passes  through  F.  Therefore  EI  HP  is  a  triangle,  and 
the  middle  points  of  EI,  EH,  and  EF  are  collinear. 

(84°,  Cor.  2) 
But  these  are  the  middle  points  of  AC,  BD,  and  EF  re- 
spectively.    .'.         P,  Q,  R  are  collinear. 

Ex.  3.  Theorem. — The  circumcentre,  the  centroid,  and  the 
orthocentre  of  a  triangle  are  col- 
linear. 

Proof.— L^\.  YD  and  ZD   be 
the  right  bisectors  of  AC  and  AB. 
Join  BY,  CZ,  and  through  E, 
A  Y  H      c  the  intersection  of  these  joins, 

draw  DE  to  meet  the  altitude  BH  in  O. 

Then  D  is  the  circumcentre  and  E  is  the  centroid.  Since 
DY  is  II  to  BH,  the  triangles  YDE  and  BOE  are  similar. 
But  BE  =  2EY,  (85°,  Cor.) 

OE  =  2DE, 
and  as  D  and  E  are  fixed  points,  O  is  a  fixed  point. 
.*.  the  remaining  altitudes  pass  through  O. 

249°.  77/^<7r^;«.— Three  concurrent  lines  perpendicular  to 
the  sides  of  a  triangle  at  X,  Y,  Z  divide  the  sides  so  that 
BX2  +  CY2  +  AZ2  =  CXHAY2-l-BZ2; 


OF  COLLINEARITY  AND  CONCURRENCE.       199 

and,  conversely,  if  three  lines  perpendicular  to  the  sides  of  a 
triangle  divide  the  sides  in  this  manner,  B 

the  lines  are  concurrent. 

Proof.— \.^\,  OX,  OY,  OZ  be  the  lines.  z^ 

Then        BX2-CX2  =  B02-C02,(i72°,i)     ^j 
Similarly  CY^- AY2  =  C02-A02, 

AZ2-BZ2  =  A02-BO^  A 

BX2  +  CY2  +  AZ2-CX"^-AY2-BZ2  =  o.  q.eul 

Conversely,  let  X,  Y,  Z  divide  the  sides  of  the  triangle  in 
the  manner  stated,  and  let  OX,  OY,  perpendiculars  to  BC 
and  CA,  meet  at  O.     Then  OZ  is  _L  to  AB. 

Proof.— 1{  possible  let  OZ'  be  J_  to  AB,  Then,  by  the 
theorem,  BXH  CY^  +  AZ'^  -  CX^  -  AY^  -  BZ'2=o, 

and  by  hyp.       BX2  +  CYHAZ2-CX2-AY2-BZ2=o, 

AZ'2-AZ2=BZ'2-BZ2. 
But  these  differences  have  opposite  signs  and  cannot  be  equal 
unless  each  is  zero.     .*.         Z'  coincides  with  Z. 

Exercises. 

1.  When  three  circles  intersect  two  and  two,  the  common 

chords  are  concurrent. 

Let  S,  Si,  S2  be  the  circles,  and  A,  B,  C  their  centres. 
Then  (113°)  the  chords  are  perpendicular  to  the  sides 
of  the  AABC  at  X,  Y,  and  Z.     And  if  r,  r^,  r^  be  the 
radii  of  the  circles, 

BX2-CX2  =  ri2-r22,  etc.,  etc., 
and  the  criterion  is  satisfied. 

.*.     the  chords  are  concurrent. 

2.  The  perpendiculars  to  the  sides  of  a  triangle  at  the  points 

of  contact  of  the  escribed  circles  are  concurrent. 

3.  When  three  circles  touch  two  and  two  the  three  common 

tangents  are  concurrent. 

4.  If  perpendiculars  from  the  vertices  of  one  triangle  on  the 

sides  of  another  be  concurrent,  then  the  perpendiculars 


200  SYNTHETIC  GEOMETRY. 

from  the  vertices  of  the  second  triangle  on  the  sides 
of  the  first  are  concurrent. 

5.  When  three  perpendiculars  to  the  sides  of  a  triangle  are 

concurrent,  the  other  three  at  the  same  distances  from 
the  middle  points  of  the  sides  are  concurrent.' 

6.  Two  perpendiculars  at  points  of  contact  of  excircles  are 

concurrent  with  a  perpendicular  at  a  point  of  contact  of 
the  incircle. 

250°.  Theorem. — When  three  points  X,  Y,  Z  lying  on  the 
sides  BC,  CA,  and  AB  of  a  triangle  are  collinear,  they  divide 
the  sides  into  parts  which  fulfil  the  relation 

f.  BX.CY.AZ     ^ 

{(X)  ■ =  I  : 

^  '  CX.AY.BZ 

and  their  joins  with  the  opposite  vertices  divide  the  angles 
into  parts  which  fulfil  the  relation 

,,N        sin  BAX .  sin  CBY .  sin  ACZ _ 

^^        sinCAX.sinABY.sinBCZ      * 

Proof  of  {a).— On  the  axis  of  X,  Y,  Z  draw  the  perpendicu- 
B  lars  AP,  BQ,  CR. 

On  account  of  similar  /\s, 

^=?S     ^=CR    AZ^AP 
CX     CR'    AY     AP'    BZ     BQ' 
BX.CY.AZ_ 
CX.AY.BZ      ' 

Proof  of  {b).-  BX=ABAX^BAsinBAX 
■^•^^^        CX     ACAX     CAsinCAX' 

sinBAX_CA    BX 
sinCAX     BA'CX' 

Similarlv        si"  CBY_AB    CY     sinACZ_BC    AZ 
^'       sinABY     CB'AY'    sin  BCZ    AC  "  BZ* 
sin  BAX  .  sin  CBY .  sin  ACZ_BX .  CY.  AZ_ 
SinCAX.sinABY.sinBCZ     CX.AY.BZ      '  q.e.d. 

The  preceding  functions  which  are  criteria  of  collinearity 


OF  COLLINEARITY  AND  CONCURRENCE.      201 
will  be  denoted  by  the  symbols 

It  is  readily  seen  that  three  points  on  the  sides  of  a  triangle 
can  be  collinear  only  when  an  even  number  of  sides  or  angles 
(2  or  o)  are  divided  internally,  and  from  230°  it  is  evident  that 
the  sign  of  the  product  is  +  in  these  two  cases. 

Hence,  in  applying  these  criteria,  the  signs  may  be  dis- 
regarded, as  the  final  sign  of  the  product  is  determined  by 
the  number  of  sides  or  angles  divided  internally. 

The  converses  of  these  criteria  are  readily  proved,  and  the 
proofs  are  left  as  an  exercise  to  the  reader. 

Ex.  If  perpendiculars  be  drawn  to  the  sides  of  a  triangle 
from  any  point  in  its  circumcircle,  the  feet  of  the  perpendicu- 
lars are  collinear. 

X,  Y,  Z  are  the  feet  of  the  perpendicu- 
lars. If  X  falls  between  B  and  C,  ^OBC 
is  <  a  ~~|,  and  therefore  Z.OAC  is  >  a  ""], 
and  Y  divides  AC  externally  ; 

.'.  it  is  a  case  of  collinearity. 

Now,  BX  =  OBcosOBC, 

and  AY=OAcosOAC. 

But,  neglecting  sign,  cosOBC=cosOAC, 

BX^OB 
AY    OA' 

CY^OC 
BZ     OB 

BX 
CX 


and  similarly. 


AZ 
CX" 


OA 
OC 


Vex/ 


and  X,  Y,  Z  are  collinear. 


Def. — The  line  of  collinearity  of  X,  Y,  Z  is  known  as 
"  Simson's  line  for  the  point  O." 


251°.   Theorem. — When  three  lines  through  the  vertices  of  a 
triangle  are  concurrent,  they  divide  the  angles  into   parts 


202  SYNTHETIC   GEOMETRY. 

which  fulfil  the  relation 

,  V        sin  BAX .  sin  CBY .  sin  ACZ_  _ 

^^'       sinCAX.sinABY.sinBCZ 
and  they  divide  the  opposite  sides  into  parts  which  fulfil  the 

^^  CX.AY.BZ 

To  prove  {a). — Let  O  be  the  point  of  concurrence  of  AX,  BY, 
and  CZ,  and  let  OP,  OQ,  OR  be  perpendiculars  on  the  sides. 

sin  BAX     OR 


Then 


sinCAX     OQ' 
sinCBY_OP 
sinABY    OR* 
sinACZ_OQ 
sinBCZ     OP' 


sinBAX\ 


A  QY 


,   .    ,     .  /sin  D/\A.\ 

.-.multiplying,  (inrCAx)=-'' 

the  negative  sign  resulting  from  the  three  angles    being 

divided  internally.  (230°) 

To  prove  (<5).— From  B  and  C  let  BE  and  CF  be  per- 
pendiculars upon  AX. 

Then,  from  similar  As  BEX  and  CFX, 
BX_BE_ABsinBAX 
CX     CF     ACsinCAX* 
e.    .,    ,  CY_BCsinCBY    AZ_CAsinACZ 

j)imiiariy,  AY ~BA sinABY'    BZ~CBsinBCZ' 

^)=-i,  from  (^).  ^^^ 

The  negative  sign  results  from  the  three  sides  being  divided 
internally.  (230°) 

It  is  readily  seen  that  three  concurrent  lines  through  the 
vertices  of  a  triangle  must  divide  an  odd  number  of  angles 
and  of  sides  internally,  and  that  the  resulting  sign  of  the 
product  is  accordingly  negative. 

Hence,  in  applying  the  criteria,  the  signs  of  the  ratios  may 
be  neglected. 


OF  COLLINEARITY  AND  CONCURRENCE.       203 

The  remarkable  relation  existing  between  the  criteria  for 
coUinearity  of  points  and  concurrence  of  lines  will  receive  an 
explanation  under  the  subject  of  Reciprocal  Polars. 

Exercises. 

252°.  I.  Equilateral  triangles  ABC,    BCA',  CAB'    are    de- 
scribed upon  the  sides  AB,  BC,  CA  of  any  triangle. 
Then  the  joins  A  A',  BB',  CC  are  concurrent. 

Proof.— S'mce       AC'  =  AB,  AB'=AC, 
and  ^CAC'=^BAB', 

ACAC'=AB'AB,  and  aAC'C  =  ^ABB'. 
i^  sin  ACZ  _sin  ACC'_AC'_AB  (oi9,°\ 

"  sin  ABY~ sin  ACC~  AC  ~ AC'  ^        ^ 

c-    .,    ,       sin  BAX     BC     sin  CBY     CA 
^'"^'^^'^^^   ihTBCZ^BA'   ilKCAX^CB' 
/sin  BAX  \_ 
VsinCAX/  ' 

and  hence  the  joins  AA',  BB',  CC  are  concurrent. 

2.  The  joins  of  the  vertices  of  a  A  with  the  points  of  con- 

tact of  the  incircle  are  concurrent. 

3.  The  joins  of  the  vertices  of  a  A  with   the   points   of 

contact  of  an  escribed  0  are  concurrent. 

4.  ABC  is  a  A,  right-angled  at  B,  CD  is  =  and  _L  to  CB, 

and  AE  is  =  and  _L  to  AB.     Then  EC  and  AD  inter- 
sect on  the  altitude  from  B. 

5.  The  internal  bisectors  of  two  angles  of  a  A  and  the 

external  bisector  of   the  third    angle    intersect    the 
opposite  sides  collinearly. 

6.  The  external  bisectors  of  the  angles  of  a  A  intersect  the 

opposite  sides  collinearly. 

7.  The  tangents  to  the  circumcircle  of  a  A,  at  the  vertices 

of  the  A,  intersect  the  opposite  sides  collinearly. 

8.  If  any  point  be  joined  to  the  vertices  of  a  A,  the  lines 

through  the  point  perpendicular  to  those  joins  intersect 
the  opposite  sides  of  the  A  collinearly. 


204 


SYNTHETIC  GEOMETRY. 


9,  A  0  cuts  the  sides  of  a  A  in  six  points  so  that  three 
of  them  connect  with  the  opposite  vertices  concurrently. 
Show  that  the  remaining  three  connect  concurrently 
with  the  opposite  vertices. 

[o.  Is  the  statement  of  Ex.  i  true  when  the  As  are  all  de- 
scribed internally  upon  the  sides  of  the  given  A  ? 

[I.  If  L  is  an  axis  of  symmetry  to  the  congruent  As  ABC 
and  A'B'C,  and  O  is  any  point  on  L,  A'O,  B'O,  and 
CO  intersect  the  sides  BC,  CA,  and  AB  collinearly. 


253°.  Theorem. — Two  triangles  which  have  their  vertices 
connecting  concurrently  have  their  corresponding  sides  inter- 
secting collinearly.     (Desargue's  Theorem.) 

\z  ABC,  A'B'C  are  two  As 
having  their  vertices  connect- 
ing concurrently  at  O,  and 
their  corresponding  sides  in- 
tersecting in  X,  Y,  Z.  To 
prove  that  X,  Y,  Z  are  col- 
linear. 

Proof. — To  the  sides  of 
AA'B'C  draw  perpendiculars 
AP,  AP',  BQ,  BQ',  CR,  CR'. 
Then,  from  similar  As, 

bx^bq; 

CX  CR' 
CY^CR' 
AY  AP' 
AZ_AF 
BZ     BQ' 

AP\  BO\CR' 

AP .  BQ .  CR* 

But  t^=«4^LAA;^;, 

sm  AA'C" 
with  similar  expressions  for  the  other  ratios. 


OF  COLLINEARITY  AND  CONCURRENCE.       205 

Also,  since  AA',  BB',  CC  are  concurrent  at  O,  they  divide 
the  angles  A'  B',  C  so  that 

sin  AA^B\  sin  BB'C.  sin  CC^A^ ^  ^ 
sin  AA'C  sin  BB'A'.  sin  CC'B'       ' 

(  — V  j  =  I,  and  X,  Y,  Z  are  collinear. 

The  converse  of  this  theorem  is  readily  proved,  and  will  be 
left  as  an  exercise  to  the  reader. 

Ex.  A',  B',  C  are  points  upon  the  sides  BC,  CA,  AB  re- 
spectively of  the  AABC,  and  AA',  BB',  CC  are  concurrent 
in  O.     Then 

1.  AB  and  A'B',  BC  and  B'C,  CA  and  CA'  meet  in  three 
points  Z,  X,  Y,  which  are  collinear. 

2.  The  lines  AX,  BY,  CZ  form  a  triangle  with  vertices  A", 
B",  C",  such  that  AA",  BB",  CC"  are  concurrent  in  O. 


OF  RECTILINEAR  FIGURES  IN  PERSPECTIVE. 

254°.  Def. — AB  and  A'B'  are  two  segments  and  AA'  and 
BB'  meet  in  O.  a  b 

Then  the  segments  AB  and  A'B'  are  said  to 
be  in  perspective  at  O,  which  is  called  their 
centre  of  perspective. 

The  term  perspective  is  introduced  from 
Optics,  because  an  eye  placed  at  O  would  see 
A'  coinciding  with  A  and  B'  with  B,  and  the 
segment  A'B'  coinciding  with  AB. 

By  an  extension  of  this  idea  O'  is  also  a 
centre  of  perspective  of  AB  and  B'A'.     O  is  o 

then  the  external  centre  of  perspective  and  O'  is  the  internal 
centre. 

Def. — Two  rectilinear  figures  of  the  same  number  of  sides 
are  in  perspective  when  every  two  corresponding  sides  have 
the  same  centre  of  perspective. 


206  SYNTHETIC   GEOMETRY. 

Cor.  I.  From  the  preceding  definition  it  follows  that  two 
rectilinear  figures  of  the  same  species  are  in  perspective  when 
the  joins  of  their  vertices,  in  pairs,  are  concurrent. 

Cor.  2.  When  two  triangles  are  in  perspective,  their  ver- 
tices connect  concurrently,  and  their  corresponding  sides 
intersect  collinearly.  (253°) 

In  triangles  either  of  the  above  conditions  is  a  criterion  of 
the  triangles  being  in  perspective. 

Def. — The  line  of  collinearity  of  the  intersections  of  corre- 
sponding sides  of  triangles  in  perspective  is  called  their  axis 
of  perspective ;  and  the  point  of  concurrence  of  the  joins  of 
corresponding  vertices  is  the  centre  of  perspective. 

255°.  Let  AA',  BB',  CC  be  six  points  which  connect  con- 
currently in  the  order  written. 

These  six  points  may  be  connected  in  four  different  ways 
so  as  to  form  pairs  of  triangles  having  the  same  centre  of 
perspective,  viz., 

ABC,  A'B'C;  ABC,  A'B'C  ;  AB'C,  A'BC;  A'BC,  AB'C. 

These  four  pairs  of  conjugate  triangles  determine  four  axes 
of  perspective,  which  intersect  in  six  points  ;  these  points  are 
centres  of  perspective  of  the  sides  of  the  two  triangles  taken 
in  pairs,  three  X,  Y,  Z  being  external  centres,  and  three  X', 
Y',  Z'  being  internal  centres.  (254°) 

The  points,  the  intersections  which  determine  them,  and 
the  segments  of  which  they  are  centres  of  perspective  are 
given  in  the  following  table  : — 


OINT. 

Determined  by 
Intersection  of 

Centre  of 
Perspective  to 

X 

BC-B'C 

BC'-B'C 

Y 

CA-C'A' 

CA'-C'A 

Z 

AB-A'B' 

AB'-A'B 

X' 

BC'-B'C 

BC-B'C 

Y' 

CA'-C'A 

CA~C'A' 

Z' 

AB'-A'B 

AB-A'B' 

OF  INVERSION  AND  INVERSE  FIGURES.       207 

And  the  six  points  lie  on  the  four  lines  thus, 
XYZ,   X'Y'Z,    X'YZ',   XY'Z'. 

Exercises 

1.  The  triangle  formed  by  joining  the  centres  of  the  three 

excircles  of  any  triangle  is  in  perspective  with  it. 

2.  The  three  chords  of  contact  of  the  excircles  of  any  tri- 

angle form  a  triangle  in  perspective  with  the  original. 

3.  The  tangents  to  the  circumcircle  of  a  triangle  at  the  three 

vertices  form  a  triangle  in  perspective  with  the  original. 


SECTION   IV. 

OF  INVERSION  AND  INVERSE  FIGURES. 

256°.  Def. — Two  points  so  situated  upon  a  centre-line  of  a 
circle  that  the  radius  is  a  geometric  mean  (169°,  Def)  between 
their  distances  from  the  centre  are  called  inverse  points  with 
respect  to  the  circle. 

Thus  P  and  Q  are  inverse  points  if 

CP.CQ  =  CB2=R2,  a| 

R  being  the  radius. 

The  0S  is  the  circle  of  inversion 
or  the  inverting  0,  and  C  is  the  centre  of  inversion. 

Cor.  From  the  definition  : — 

1.  An  indefinite  number  of  pairs  of  inverse  points  may  lie  on 

the  same  centre-line. 

2.  An  indefinite  number  of  circles  may  have  the  same  two 

points  as  inverse  points. 

3.  Both  points  of  a  pair  of  inverse  points  lie  upon  the  same 

side  of  the  centre  of  inversion. 

4.  Of  a  pair  of  inverse  points  one  lies  within  the  circle  and 

one  without. 


208  SYNTHETIC  GEOMETRY. 

5.  P  and  Q  come  together  at  B  ;  so  that  any  point  on  the 

circle  of  inversion  is  its  own  inverse. 

6.  When  P  comes  to  C,  Q  goes  to  co  ;  so  that  the  inverse  of 

the  centre  of  inversion  is  any  point  at  infinity. 

257°.  Problem. — To  find  the  circle  to  which  two  pairs  of 
coUinear  points  may  be  inverse  points. 

P^,_-— .^Q   c    p! ~^Q'  P>  Qj  P')  Q'  ^6  the  four 

collinear  points,  of  which 
PQ  and  P'Q'  are  respec- 
tively to  be  pairs  of  in- 
verse points. 

Through     P,     Q     and 
\v  through    P',   Q'   describe 

any  two  circles  S,  S'  to  intersect  in  two  points  U  and  V.  Let 
the  connector  UV  cut  the  axis  of  the  points  in  C,  and  let 
CT  be  a  tangent  to  circle  S'.  Then  C  is  the  centre  and  CT 
the  radius  of  the  required  circle. 

Proof.—     CT2  =  CP'.  CQ'  =  CU.  CV  =  CQ.  CP. 

Cor.  If  the  points  have  the  order  P,  P',  Q,  Q'  the  centre  C 
is  real  and  can  be  found  as  before,  but  it  then  lies  within 
both  circles  S  and  S',  and  no  tangent  can  be  drawn  to 
either  of  these  circles  ;  in  this  case  we  say  that  the  radius 
of  the  circle  is  imaginary  although  its  centre  is  real. 
In  the  present  case  P  and  Q,  as  also  P'  and  Q',  lie  upon 
opposite  sides  of  C,  and  the  rectangles  CP.  CQ  and  CP'.  CQ' 
are  both  negative.  But  R^  being  alwa;^s  positive  (163°) 
cannot  be  equal  to  a  negative  magnitude. 

When  the  points  have  the  order  P,  P',  Q',  Q,  the  circle  of 
inversion  is  again  real. 

Hence,  in  order  that  the  circle  of  inversion  may  be  real, 
each  pair  of  points  must  lie  wholly  without  the  other,  or  one 
pair  must  lie  between  the  others. 


OF  INVERSION  AND  INVERSE  FIGURES.       209 

Exercises. 

1.  Given  a  0  and  a  point  without  it  to  find  the  inverse  point. 

2.  Given  a  0  and  a  point  within  it  to  find  the  inverse  point. 

3.  Given  two  points  to  find  any  0  to  which  they  shall  be 

inverse. 

4.  In  3  the  0  is  to  have  a  given  radius. 

5.  In  3  the  0  is  to  have  a  given  centre  on  the  line  of  the 

points. 

258°.  Theorem. — A  0  which  passes  through  a  pair  of 
inverse  points  with  respect  to  another  0  cuts  the  latter 
orthogonally.  (i  15°,  Defs.  i,  2) 

And,  conversel}^,  a  0  which  cuts  another  0  orthogonally 
determines  a  pair  of  inverse  points  on  any  centre-line  of  the 
latter. 

1.  P  and  O  are  inverse  to  08. 
Then        CP.C0  =  CT2, 

CT  is  tangent  to  0S'. 

(176°,  Cor.  3) 
And  .*.  S'  cuts  S  orthogonally 
since  the  radius  of  S  is  perpen- 
dicular to  the  radius  of  S'  at  its  end-point. 

2.  Conversely,  let  S'  cut  S  orthogonally.  Then  Z-CTC 
is  a  ""I,  and  therefore  CT  is  tangent  to  S'  at  the  point  T. 
Hence  CT^^CP.CQ, 

and  P  and  Q  are  inverse  points  to  08. 

Cor.  I.  A  0  through  a  pair  of  points  inverse  to  one  another 
with  respect  to  two  0s  cuts  both  orthogonally. 

Cor.  2.  A  0  which  cuts  two  0s  orthogonally  determines 
on  their  common  centre-line  a  pair  of  points  which  are  inverse 
to  one  another  with  respect  to  both  0s. 

Cor.  3.  If  the  08  cuts  the  0s  8'  and  8"  orthogonally,  the 
tangents  from  the  centre  of  8  to  the  0s  8'  and  8"  are  radii  of 
S  and  therefore  equal. 


2IO 


SYNTHETIC   GEOMETRY. 


.*.  (178°)  a  0  which  cuts  two  0s  orthogonally  has  its 
centre  on  their  radical  axis. 

Cor.  4,  A  0  having  its  centre  on  the  radical  axis  of  two 
given  0s,  and  cutting  one  of  them  orthogonally,  cuts  the 
other  orthogonally  also. 


259°.  Let  P,  O  be  inverse  points  to  circle  S  and  D  any 

point  on  it. 
Then 
CP.CQ=CD2, 
.-.    CP:CD  =  CD:CQ. 
Hence  the  triangles  CPD 
and  CDQ  are  similar,  and 
PD  and  DQ  are  homolo- 
gous sides. 
PD2_CD2_CP.CQ_CP 
QD2     CQ2        CQ2        CQ- 
.'.  the  squares  on  the  joins  of  any  point  on  a  circle  with  a 
pair  of  inverse  points  with  respect  to  the  circle  are  propor- 
tional to  the  distances  of  the  inverse  points  from  the  centre. 

Cor.  I.  If  P  and  Q  are  fixed,  PD^:  QD^  is  a  fixed  ratio. 
.'.  the  locus  of  a  point,  for  which  the  squares  on  its  joins  to 
two  fixed  points  have  a  constant  ratio,  is  a  circle  having  the 
two  fixed  points  as  inverse  points. 

Cor.  2.  When  D  comes  to  A  and  B  we  obtain 

CP_PD2_PA2_PB2 

CQ     QD'-*     0A2     QB2' 

Pp^PA^PB 

QD     QA     QB- 
Hence  DA  and  DB  are  the  bisectors  of  the  ^PDQ,  and 
the  segments  PB  and  BQ  subtend  equal  angles  at  D. 

Hence  the  locus  of  a  point  at  which  two  adjacent  segments 
of  the  same  line  subtend  equal  angles  is  a  circle  passing 
through  the  common  end-points  of  the  segments  and  having 
their  other  end-points  as  inverse  points. 


OF  INVERSION  AND  INVERSP:  FIGURES.      211 

Cor.  3.  Let  P',  Q'  be  a  second  pair  of  inverse  points.    Then 
Z.BDP'  =  ^BDQ',  and  ^BDP  =  ^BDQ, 
z.PDP'  =  ^QDQ'; 
or  the  segments  PP'  and  QO'  subtend  equal  angles  at  D. 

Hence  the  locus  of  a  point  at  which  two  non-adjacent  seg- 
ments of  the  same  line  subtend  equal  angles  is  a  circle  having 
the  end-points  as  pairs  of  inverse  points. 

Cor.  4.  Since        AP  :  AQ  =  PB  :  BQ,  (Cor.  i) 

.*.  P  and  Q  divide  the  diameter  AB  in  the  same  manner  in- 
ternally and  externally,  and  B  and  A  divide  the  segment  PQ 
in  the  same  manner  internally  and  externally. 

.*.  from  208°,  Cor.,  P,  Q  divide  AB  harmonically,  and  A,  B 
divide  PO  harmonically. 

Hence,  when  two  segments  of  the  same  line  are  such  that 
the  end-points  of  one  divide  the  other  harmonically,  the  circle 
on  either  segment  as  diameter  has  the  end-points  of  the  other 
segment  as  inverse  points. 


Exercises. 

1.  If  a  variable  circle  passes  through  a  pair  of  inverse  points 

with  respect  to  a  fixed  circle,  the  common  chord  of  the 
circles  passes  through  a  fixed  point. 

2.  To  draw  a  circle  so  as  to  pass  through  a  given  point  and 

cut  a  given  circle  orthogonally. 

3.  To  draw  a  circle  to  cut  two  given  circles  orthogonally. 

4.  On  the  common  centre-line  of  two  circles  to  find  a  pair  of 

points  which  are  inverse  to  both  circles. 

Let  C,  C  be  the  centres  of  the  circles  S  and  S'.  Take 
any  point  P,  without  both  circles,"  and  find  its  inverses 
P'  and  P"  with  respect  to  both  circles.         (257°,  Ex.  i) 

The  circle  through  P,  P',  and  P"  cuts  the  common 
centre-line  CC  in  the  required  points  Q  and  Q'. 

5.  To  describe  a  circle  to  pass  through  a  given  point  and  cut 

two  given  circles  orthogonally. 


212  SYNTHETIC   GEOMETRY. 

6.  To  determine,  on  a  given  line,  a  point  the  ratio  of  whose 

distances  from  two  fixed  points  is  given. 

7.  To  find  a  point  upon  a  given  line  from  which  the  parts  of 

a  given  divided  segment  may  subtend  equal  angles. 

8.  A,  B,  C,  D  are  the  vertices  of  a  quadrangle.     Gn  the 

diagonal  BD  find  a  point  at  which  the  sides  BA  and 
BC  subtend  equal  angles. 

9.  To  draw  a  circle  to  pass  through  a  given  point  and  touch 

two  given  lines. 

260°.  Def. — One  figure  is  the  inverse  of  another  when  every 
point  on  one  figure  has  its  inverse  upon  the  other  figure. 

Theorem. — The  inverse  of  a  circle  is  a  circle  when  the 
centre  of  inversion  is  not  on  the  figure  to  be  inverted. 

s 


T 


Let  O  be  the  centre  of  inversion  and  S  be  the  circle  to  be 
inverted  ;  and  let  A',  B',  C  be  the  inverses  of  A,  B,  C  respec- 
tively.    To  prove  that  the  L^Ch!  =  ~^ 

Pr^t?/— OA.  OA'=OB.  OB'  =  OC.  0C'=R2,  (256°) 

A0A'C5=?A0CA,  and  AOB'C'^AOCB. 
(i)  aOC'A'=z.OAC,  and  (2)  Z-OC'B'  =  z.OBC. 
And  ^B'CA'  =  ^OCA'-OCB'=^OAC-^OBC 

=  z.ACB=n, 
since  ACB  is  in  a  semicircle. 

.*.  as  C  describes  S,  its  inverse,  C,  describes  the  circle  S' 
on  A'B'  as  diameter.  q.e.d. 


OF  INVERSION  AND  INVERSE  FIGURES.      21  3 

Cor.  I.  The  point  O  is  the  intersection  of  common  direct 
tangents. 

Cor.  2.  •.•  CD'.  OD  =  OT.  OT'.  ^^.^  =  RK  ?-^', 

where  R  is  the  radius  of  the  circle  of  inversion  ; 
.*.  the  centre  of  a  circle  and  the  centre  of  its  inverse  are  not 
inverse  points,  unless  OD  =  OT,  i.e.,  unless  the  centre  of 
inversion  is  at  00 . 

Cor.  3.  When  the  circle  to  be  inverted  cuts  the  circle  of 
inversion,  its  inverse  cuts  the  circle  of  inversion  in  the  same 
points.  (256°,  Cor.  5) 

261°.  Theorem. — A  circle  which  passes  through  the  centre 
of  inversion  inverts  into  a  line. 

Let  O  be  the  centre  of  inversion 
and  S  the  circle  to  be  inverted,  and 
let  P  and  P'  be  inverses  of  Q  and  Q'. 

Proof.— '$AnQ.Q 

OP.OO  =  OP'.  00'  =  R2, 
OP:OP'  =  OQ':OQ, 
and  the  triangles  OPP'  and  OQ'Q 
are  similar,  and  ^OPP'  =  Z.OQ'Q  =  "~|>  since  OQ'Q  is  in  a 
semicircle.  And  as  this  is  true  however  OP'  be  drawn,  PP' 
is  a  line  ±  to  OP,  the  common  centre-line  of  the  circle  of 
inversion  and  the  circle  to  be  inverted.  q.e.d. 

Cor.  I.  Since  inversion  is  a  reciprocal  process,  the 
inverse  of  a  line  is  a  circle  through  the  centre  of  inversion 
and  so  situated  that  the  line  is  J_  to  the  common  centre- 
line of  the  two  circles. 

Cor.  2.  Let  I  be  the  circle  of  inversion,  and  let  PT  and  PT' 
be  tangents  to  circles  I  and  S  respectively.     Then, 

PT2  =  OP2-OT2=OP2-OP.OQ  =  OP.PQ=PT'2, 
PT  =  Pr, 
.*.  when  a  circle  inverts  into  a  line  with  respect  to  another 
circle,  the  line  is  the  radical  axis  of  the  two  circles.  (178°,  Def.) 


214 


SYNTHETIC   GEOMETRY. 


Cor.  3.  If  a  circle  passes  through  the  centre  of  inversion 
and  cuts  the  circle  of  inversion,  its  inverse  is  their  common 
chord. 

Cor.  4.  A  centre-line  is  its  own  inverse. 

Cor.  5.  Considering  the  centre  of  inversion  as  a  point-circle, 
its  inverse  is  the  line  at  00 . 


262°.  A  circle  which  cuts  the  circle  of  inversion  orthogon- 
ally inverts  into  itself. 

Since  circle  S  cuts  circle  I  orthogon- 
ally OT  is  a  tangent  to  S,  and  hence 

OP.OO  =  OT2, 
.'.  P  inverts  into  Q  and  O  into  P,  and 
the  arc  TQV  inverts  into  TPV  and  vice 
versa.  q.e.d. 

Cor.  Since  I  cuts  S  orthogonally,  it  is  evident  that  I  inverts 
into  itself  with  respect  to  S. 


263°.  A  circle,  its  inverse,  and  the  circle  of  inversion  have 
t^       a  common  radical  axis. 

Let  I  be  the  circle  of  inver- 
'^^^^^     I  sion,  and  let  the  circle  S'  be 

the  inverse  of  S. 
o^---^  \J  I    \  The  tangents  TT'  and  VV 

v^^^-~f-f_  X  meet  at  O  (260°,  Cor.  i),  and 

T,  T'  are  inverse  points.  D, 
the  middle  point  of  TT'  is  on  the  radical  axis  of  S  and  S', 
and  the  circle  with  centre  at  D  and  radius  DT  cuts  S  and  S' 
orthogonally.  But  this  circle  also  cuts  circle  I  orthogonally 
(258°).     .*.  D  is  on  the  radical  axes  of  I,  S  and  S'. 

Similarly  D',  the  middle  point  of  the  tangent  VV,  is  on  the 
radical  axes  of  I,  S  and  S'. 

.*.  the  three  circles  I,  S,  and  S'  have  a  common  radical 
axis  passing  through  D  and  D'.  q.e.d. 


OF  INVERSION  AND  INVERSE  FIGURES.      21  5 

Remarks. — This  is  proved  more  simply  by  supposing  one 
of  the  circles  to  cut  the  circle  of  inversion.  Then  its  inverse 
must  cut  the  circle  of  inversion  in  the  same  points,  and  the 
common  chord  is  the  common  radical  axis. 

The  extension  to  cases  of  non-intersection  follows  from  the 
law  of  continuity. 

264°.  Theorem. — The  angle  of  intersection  of  two  lines  or 
circles  is  not  changed  in  magnitude  by  inversion. 

Let  O  be  the  centre  of  in- 
version, and  let  P  be  the  point 
of  intersection  of  two  circles  S 
and  S',  and  Q  its  inverse. 
Take  R  and  T  points  near  P,  o 
and  let  U  and  V  be  their  in- 
verses.    Then 

0U.0R  =  0Q.0P  =  0V.0T  =  R2, 
AOQU^AORP, 
and  ^OQU  =  ^ORP  =  ^RPX-^ROP. 

Similarly        ^OQV  =  ^TPX-^TOP, 
^UQV  =  ^RPT-^ROT. 
But  at  the  limit  when  R  and  T  come  to  P  the  angle  between 
the  chords  RP  and  PT  becomes  the  angle  between  the  circles 
(115°,  Def.  I ;   109°,  Def.  i).    And,  since  ^ROT  then  vanishes, 
we  have  ultimately        ^UQV  =  I.RPT. 

Therefore  S  and  S',  and  their  inverses  Z  and  Z',  intersect  at 
the  same  angle. 

Cor.  I.  If  two  circles  or  a  line  and  a  circle  touch  one 
another  their  inverses  also  touch  one  another. 

Cor.  2.  If  a  circle  inverts  into  a  line,  its  centre-lines  invert 
into  circles  having  that  line  as  a  common  diameter.  For, 
since  the  circle  cuts  its  centre-lines  orthogonally,  their  in- 
verses must  cut  orthogonally.  But  the  centre-line  is  the  only 
line  cutting  a  circle  orthogonally. 


2l6 


SYNTHETIC   GEOMETRY. 


Exercises. 

1.  What  is  the  result  of  inverting  a  triangle  with  respect  to 

its  incircle  ? 

2.  The  circle  of  self-inversion  of  a  given  circle  cuts  it  or- 

thogonally. 

3.  Two  circles  intersect  in  P  and  O,  and  AB  is  their  common 

centre-line.     What  relation  holds  between  the  various 
parts  when  inverted  with  P  as  centre  of  inversion  ? 

4.  A  circle  cuts  two  circles  orthogonally.     Invert  the  system 

into  two  circles  and  their  common  centre-line. 

5.  Three  circles  cut  each  other  orthogonally.     If  two  be  in- 

verted into  lines,  their  intersection  is  the  centre  of  the 
third. 


265°.  The  two  following  examples  are  important. 

Ex.  I.  Any  two  circles  cut  their  common  centre-line,  and  a 
circle  which  cuts  them  orthogonally  in  two  sets  of  points 
which  connect  concurrently  on  the  last-named  circle. 

S  and  S'  are  the 
two  given  circles  and 
Z  a  circle  cutting 
them  orthogonally. 

Invert  S  and  S' 
and  their  common 
centre  line  with  re- 
spect to  a  circle 
which  cuts  S  and  S'  orthogonally  and  has  its  centre  at  some 
point  O  on  Z.  S  and  S'  invert  into  themselves,  and  their 
centre-line  into  a  circle  through  O  cutting  S  and  S'  ortho- 
gonally, i.e.^  into  circle  Z. 

.•.  A'  is  the  inverse  of  A,  B'  of  B,  etc  ,  and  the  points  AA', 
BB',  CC,  DD'  connect  concurrently  at  O. 

Ex.  2.  The  nine-points  circle  of  a  triangle  touches  the  in- 
circle and  the  excircles  of  the  triangle. 


OF  INVERSION  AND  INVERSE  FIGURES.     21/ 

Let  ABC  be  any  triangle  having  its  side  AB  touched  by 

the    incircle    I    at     b  h  ^ 

T,  and  by  the  ex- 
circle  to  the  side 
c  at  T'.  Take 
CH  =  CAand  CD 
=  CB,  and  join  DH 
and  HA. 

From  the  sym- 
metry of  the  fig- 
ure it  is  evident 
that  HD  touches 
both  the  circles  I  and  S.  Let  E  and  F  be  the  middle  points 
of  AB  and  AC,  and  let  EF  cut  HA  in  G. 

From  135°,  Ex.  i,  AT  =  BT'=j-rt;, 

whence  ET  =  ET'  =  |(^-/^). 

But,  since  EF  bisects  HA,  EG  =  ^BH  =  |(^-^), 

ET  =  ET'  =  EG, 
and  the  circle  with  E  as  centre  and  EG  as  radius  cuts  I  and 
S  orthogonally,  and,  with  respect  to  this  circle,  the  circles  I 
and  8  invert  into  themselves. 

Now,     PF  :HC  =  DF  :DC  =  BC-CF:BC, 

PF  =  ii^(BC-CF)  =  ^-— , 
BC  2a 

EP  =  EF-PF  =  --^-j-^, 
2  2a 

EP.EF  =  f''-^-h— )-  =  K^-^)2  =  EG2. 

^2  2a'2 

.'.  P  inverts  into  F,  and  the  line  HD  into  the  circle  through 
E  and  F,  and  by  symmetry,  through  the  middle  point  of  BC. 
But  this  is  the  nine-points  circle  (116°,  Ex.  6).  And  since 
HD  touches  I  and  S,  the  nine-points  circle,  which  is  the 
inverse  of  HD,  touches  the  inverses  of  I  and  S,  t.e.,  I  and  S 
themselves. 

And,   similarly,   the   nine-points   circle    touches    the   two 
remaining  excircles. 


2l8  SYNTHETIC   GEOMETRY. 


SECTION  V. 

OF    POLE   AND    POLAR. 

266°.  Def. — The  line  through  one  of  a  pair  of  inverse  points 
perpendicular  to  their  axis  is  the  polat'  of  the  other  point  with 
respect  to  the  circle  of  inversion,  and  the  point  is  the  pole  of 
the  line. 

The  circle  is  called,  in  this  relation,  Xh^  polar  circle,  and  its 
centre  is  called  the  polar  cejitre. 

From  this  definition  and  from  the  nature  of  inverse  points 
we  readily  obtain  the  following  : — 

Cor.  I.  The  polar  of  the  polar  centre  is  a  line  at  infinity. 
But,  since  the  point  which  is  the  inverse  of  the  centre  may  go 
to  00  along  any  centre-line,  all  the  hnes  obtained  therefrom 
are  polars  of  the  centre.  And  as  a  point  has  in  general  but 
o/te  polar  with  respect  to  any  o?ie  circle,  we  speak  of  the  polar 
of  the  centre  as  being  the  line  at  infinity,  thus  assuming  that 
there  is  but  one  line  at  infinity. 

Cor.  2.  The  polar  of  any  point  on  the  circle  is  the  tangent 
at  that  point ;  or,  a  tangent  to  the  polar  circle  is  the  polar  to 
the  point  of  contact. 

Cor.  3.  The  pole  of  any  line  lies  on  that  centre-line  of  the 
polar  circle  which  is  perpendicular  to  the  former  line. 

Cor.  4.  The  pole  of  a  centre-line  of  the  polar  circle  lies 
at  00  on  the  centre-line  which  is  perpendicular  to  the 
former. 

Cor.  5.  The  angle  between  the  polars  of  two  points  is 
equal  to  the  angle  subtended  by  these  points  at  the  polar 
centre. 

267°.  Theorem. — If  P  and  Q  be  two  points,  and  P  lies  on 
the  polar  of  Q,  then  O  lies  on  the  polar  of  P. 


OF   POLE   AND   POLAR.  219 

OP    and    OQ    are    centre-lines    of   the    polar    circle 
and  PE,  ±  to  OQ,  is  the  polar  of  Q. 
To  prove  that  QD,  ±  to  OP,  is  the 
polar  of  P. 

Proof.— The  As  ODQ  and  OEP 
are  similar. 

OE:OP  =  OD:OQ, 
and.-.         OE.OQ  =  OP.OD. 
But  E  and  Q  are  inverse  points  with 
respect  to  circle  I, 

P  and  D  are  inverse  points, 
and  .'.  DQ  is  the  polar  of  P. 

Def. — Points  so  related  in  position  that  each  lies  upon  the 
polar  of  the  other  are  conjugate  points,  and  lines  so  related 
that  each  passes  through  the  pole  of  the  other  are  conjugate 
lines. 

Thus  P  and  Q  are  conjugate  points  and  L  and  M  are  con- 
jugate lines. 

Cor.  I.  If  Q  and,  accordingly,  its  polar  PV  remain  fixed 
while  P  moves-  along  PE,  L,  which  is  the  polar  of  P,  will 
rotate  about  Q,  becoming  tangent  to  the  circle  when  P  comes 
to  U  or  V,  and  cutting  the  circle  when  P  passes  without. 

Similarly,  if  Q  moves  along  L,  M  will  rotate  about  the 
point  P. 

Cor.  2.  As  L  will  touch  the  circle  at  U  and  V,  UV  is  the 
chord  of  contact  for  the  point  Q. 

.*.  for  any  point  without  a  circle  its  chord  of  contact  is  its 
polar. 

Cor.  3.  For  every  position  of  P  on  the  line  M,  its  polar 
passes  through  Q. 

.'.  collinear  points  have  their  polars  concurrent,  and  con- 
current lines  have  their  poles  collinear,  the  point  of  concur- 
rence being  the  pole  of  the  line  of  collinearity. 


220  SYNTHETIC   GEOMETKY. 

Exercises. 

1.  Given  a  point  and  a  line  to  find  a  circle  to  which  they  are 

pole  and  polar. 

2.  In  Ex,  I  the  circle  is  to  pass  through  a  given  point. 

3.  In  the  figure  of  267°  trace  the  changes, 

(a)  when  P  goes  to  00  along  M  ; 

(6)  when  P  goes  to  00  along  OD  ; 

(c)  when  P  moves  along  UV,  what  is  the  locus  of  D  ? 

4.  From  any  point  on  a  circle  any  number  of  chords  are 

drawn,  show  that  their  poles  all  lie  on  the  tangent  at 
the  point. 

5.  On  a  tangent  to  a  circle  any  number  of  points  are  taken, 

show  that  all  their  polars  with  respect  to  the  circle  pass 
through  the  point  of  contact. 


268°.   Theorem. — The  point  of  intersection  of  the  polars  of 
two  points  is  the  pole  of  the  join  of  the  points. 

A  Let  the  polars  of  B  and  of  C  pass  through  A. 

Then  A  lies  on  the  polar  of  B,  and  therefore  B 

lies   on   the  polar  of  A  (267°).      For  similar 

B .  .  c  reasons  C  lies  on  the  polar  of  A. 

.*.  the  polar  of  A  passes  through  B  and  C  and  is  their  join. 

q.e.d. 

Cor.  Let  two  polygons  ABCD...  and  abc...  be  so  situated 
that  a  is  the  pole  of  AB,  b  of  BC,  c 
of  CD,  etc. 

Then,  since  the  polars  of  a  and  b 
meet  at  B,  B  is  the  pole  of  ab  ; 
similarly  C  is  the  pole  of  hc^  etc. 
kE  .'.if  two  polygons  are  such  that 
the  vertices  of  one  are  poles  of  the  sides  of  the  other,  then, 
reciprocally,  the  vertices  of  the  second  polygon  are  poles  of 
the  sides  of  the  first,  the  polar  circle  being  the  same  in 
each  case. 


OF   POLE   AND   POLAR. 


221 


Def.  I.- -Polygons  related  as  in  the  preceding  corollary  are 
polar  reciprocals  to  one  another. 

Def.  2. — When  two  polar  reciprocal  As  become  coincident, 
the  resulting  A  is  self-reciprocal  or  self-conjngate^  each  vertex 
being  the  pole  of  the  opposite  side. 

I^^f'  3- — The  centre  of  the  0  with  respect  to  which  a  A  is 
self-reciprocal  is  the  polar  centre  of  the  A?  and  the  O  itself 
is  the  polar  circle  of  the  A 


269°.  The  orthocentre  of  a  triangle  is  its  polar  centre 
.  Let  ABC   be  a  self-conjugate  A-  a 
Then  A  is  the  pole  of  BC,  and  B  of 
AC,  and  C  of  AB. 

Let  AX,  J_  to  BC,  and  BY,  ±  to 
AC,  meet  in  O.  Then  O  is  the  ortho- 
centre.  (88°,  Def) 

Now,  as  AX  is  \_  to  BC,  and  as  A 
is  the  pole  of  BC,  the  polar  centre 
lies  on  AX.  For  similar  reasons  it 
lies  on  BY.  (266°,  Cor.  3) 

.*.  O  is  the  polar  centre  of  the  AABC.  q.e.d. 

Cor.  I.  With  respect  to  the  polar  0  of  the  A?  the  0  on 
AO  as  diameter  inverts  into  a  line  _L  to  AO  (261°).  And  as 
A  and  X  are  inverse  points,  this  line  passes  through  X  ; 
therefore  BC  is  the  inverse  of  the  0  on  OA  as  diameter. 

Similarly,  AC  is  the  inverse  of  the  0  on  OB  as  diameter, 
and  AB  of  the  0  on  OC  as  diameter. 

Cor.  2.  As  the  0  on  AO  inverts  into  BC,  the  point  D  is 
inverse  to  itself,  and  is  on  the  polar  0  of  the  A-         (256°,  5) 

.'.  OD  is  the  polar  radius  of  the  A- 

Cor.  3.  If  O  falls  within  the  A^  it  is  evident  that  the  0  on 
OA  as  diameter  will  not  cut  CB.  In  this  case  the  polar 
centre  is  real  while  the  polar  radius  is  imaginary.  (257°,  Cor.) 

Hence  a  A  which  has  a  real  polar  circle  must  be  obtuse- 
angled. 


222  SYNTHETIC   GEOMETRY. 

Cor.  4.  The  0  on  BC  as  diameter  passes  through  Y  since 

Y  is  a  ~~1- 

But  B  and  Y  are  inverse  points  to  the  polar  0. 

.*.  the  polar  0  cuts  orthogonally  the  0  on  BC  as  diameter. 

(258°) 

Similarly  for  the  circles  on  CA  and  AB. 

.'.  the  polar  0  of  a  A  cuts  orthogonally  the  circles  having 
the  three  sides  as  diameters. 

Cor.  5.  The^A0Z=^B,^BOZ  =  zA,  and  ^OAC=^(C-|)- 
And  CX  =  OC  sin  AOZ  =  OC  sin  B,  also  =-^cosC, 

0C=--^ — —  .cosC= -^cos  C, 
sm  B 

where  d  is  the  diameter  of  the  circumcircle  (228°)  to  the 

triangles  AOC  or  BOC  or  AOB  or  ABC,  these  being  all 

equal.  (116°,  Ex.  4) 

Similarly         OA  =  ^cosA,  OB  =  ^cosB. 
But  OX  =  OCcosB=-^cosBcosC, 

R2  =  OX.  OA=-^2cosAcosBcosC. 

In  order  that  the  right-hand  member  may  be  +,  one  of  the 
angles  must  be  obtuse. 

Cor.  6.  R2  =  OC.OZ  =  OC(OC  +  CZ)  =  OC2  +  OC.CZ, 
and  OC=-<^cosC,  and  CZ  =  ^sinB  =  ^,  (228°) 

R2=^2(i_sin2C)-rt^cosC 

=  ^2_^(^2  +  ^2  +  ^2).  (217°) 

If  O  is  within  the  triangle,  d"^  <  \{a^ ^  b'^ ■\- c^)  and  R  is 
imaginary. 

Exercises. 

1.  If  two  triangles  be  polar  reciprocals,  the  inverse  of  a  side 

of  one  passes  through  a  vertex  of  the  other. 

2.  A  right-angled  triangle  has  its  right-angled  vertex  at  the 

centre  of  a  polar  circle.     What  is  its  polar  reciprocal  .'* 

3.  In  Fig.  of  269°,  if  the  polar  circle  cuts  CY  produced  in  C, 

prove  that  CY=YC'. 


OF   POLE   AND   POLAR.  223 

4.  If  P  be  any  point,  ABC  a  triangle,  and  A'B'C  its  polar 
reciprocal  with  respect  to  a  polar  centre  O,  the  per- 
pendiculars from  O  on  the  joins  PA,  PB,  and  PC 
intersect  the  sides  of  A'B'C  coUinearly. 

270°.  Theore77t.—\i  two  circles  intersect  orthogonally,  the 
end-points  of  any  diameter  of  either  are  conjugate  points 
with  respect  to  the  other. 

Let  the  circles  S  and  S '  in- 
tersect orthogonally,  and  let 
PQ  be  a  diameter  of  circle  S'. 
Then  P'  is  inverse  to  P,  and 
P'Q  is  ±  to  CP. 

.'.  P'Q  is  the  polar  of  P  with 
respect  to  circle  S. 

.'.  Q  lies  on  the  polar  of  P,  and  hence  P  lies  on  the  polar 
of  O,  and  P  and  Q  are  conjugate  points  (267°  and   Def.). 

q.e.d. 

Cor.  I.  PQ2  =  CP2  +  CQ2-2CP.CP'  (172°,  2) 

=  CP2  +  C02-2R2 

=  CP2-R2  +  CQ2-R2  =  T2  +  T'2, 
where  T  and  T'  are  tangents  from  P  and  Q  to  the  circle  S. 

.*.  the  square  on  the  join  of  two  conjugate  points  is  equal 
to  the  sum  of  the  squares  on  the  tangents  from  these  points 
to  the  polar  circle. 

Cor.  2.  If  a  circle  be  orthogonal  to  any  number  of  other 
circles,  the  end-points  of  any  diameter  of  the  first  are  conju- 
gate points  with  respect  to  all  the  others.  And  when  two 
points  are  conjugate  to  a  number  of  circles  the  polars  of  either 
point  with  respect  to  all  the  circles  pass  through  the  other 
point. 

271°.  Theorhn. — The  distances  of  any  two  points  from  a 
polar  centre  are  proportional  to  the  distances  of  each  point 
from  the  polar  of  the  other  with  respect  to  that  centre. 

(Salmon) 


224 


SYNTHETIC   GEOMETRY. 


NN'  is  the  polar  of  P  and  MM'  is  the  polar  of  Q. 

PO:00  =  PM:ON. 


Then 


MM' 


Proof.— h^t  Om  be  1|  to 
and  O;/ be  II  to  NN'.     Then 
OP'.  OP  =  OQ'.  OQ 
OP_OQ'_M;;^ 
OQ     OP'     N;^" 
But  the  triangles  OP;;z  and  00;/ 
are  similar, 

OQ     Q«      N;2 

_P^+M;;g_PM  , 

Q;/  +  N«      ON"    ^ 

Cor.  I.  A,  B  are  any  two  points  and  L  and  M  their  polars, 
and  P  the  point  of  contact  of  any  tangent  N. 

AX  and  BY  are  _L  upon  N,  and  PH  and  PK  are  J_  upon  L 
and  M  respectively.     Then 

M  =  ?2and  AX_AO 
PK      R  PH      R' 

BY. AX     AO.BO     ;_ ^,^„, 

PK.-pH  =  -^^-=^'^^°^^''"'' 
AX.BY  =  ^.PH.PK. 
If  A  and  B  are  on  the  circle,  L  and  M  become  tangents 
having  A  and  B  as  points  of  contact,  and  AO  =  BO  =  R. 

AX.BY  =  PH.PK.  (See  211°,  Ex.  i) 


Exercises. 

If  P  and  Q  be  the  end-points  of  any  diameter  of  the  polar 
circle  of  the  AABC,  the  chords  of  contact  of  the  point 
P  with  respect  to  the  circles  on  AB,  BC,  and  CA  as 
diameters  all  pass  through  Q. 

Two  polar  reciprocal  triangles  have  their  corresponding 
vertices  joined.  Of  what  points  are  these  joins  the 
polars  ? 


OF   POLE  AND   POLAR. 


225 


A,  B,  C  are  the  vertices  of  a  triangle  and  L,  M,  N  the 
corresponding  sides  of  its  reciprocal  polar.     If  T  be  a 
tangent  at  any  point  P,  and  AT  is  ±  to  T,  etc., 
AT.BT.CT     AG. BO. CO 

PL.PM.PN  = R3 =^  '^"''^^'- 

If  A,  B,  C  are  on  the  circle, 

AT .  BT .  CT  =  PL .  PM  .  PN. 
In  Ex.  3,  if  A',  B',  C  be  the  vertices  of  the  polar  reciprocal, 
AT .  BT .  CT_  A^O  .  B^O  .  CO 
AT.BT.CT  R3 

The  right-hand  expression  is  independent  of  the  position 
ofT. 
If  ABC,  A'B'C  be  polar  reciprocal  triangles  whose  sides 
are  respectively  L,  M,  N  and  L',  M',  N',  and  if  AM'  is 
the  ±  from  A  to  M',  etc., 

AM'.  BN'.  CL'  =  AN'.  BL'.  CM', 
and         A'M .  B'N  .  C'L  =  A'N .  B'L .  CM. 


etc., 


272°.   Theorem. — Triangles  which  are  polar  reciprocals  to 
one  another  are  in  perspective.  ^' 

Let  ABC  and  A'B'C  be  polar  recipro- 
cals.    Let  AP,  AP'  be  perpendiculars  on  ^ 
A'B'  and  A'C,  BQ  and  BO'  be  perpen- 
diculars on  B'C  and  B'A',  etc. 

Then  (271°)  AP'_AO     BQ'^BO 
men  (271)  ^--^,    CR     CO 

AP'.  BQ'.  CR'_ 
AP.B(2.CR       ' 
But     AP'  =  AA'sinAA'P',   AP  =  AA'sin  AA'P',     . 
.    AP'^sinAA'P' 
AP     sin  AA'P' 
and  similarly  for  the  other  ratios.     Hence  AA',  BB',  CC 
divide  the  angles  at  A,  B,  and  C,  so  as  to  fulfil  the  criterion 
of  251°. 

.*.  AA',  BB',  and  CC  are  concurrent,  and  the  triangles  are 
in  perspective.  (254°,  Cor.  2) 

P 


226  SYiNTHETIC   GEOMETRY. 

SECTION    VI. 

OF   THE   RADICAL  AXIS. 

273°.  Dcf.  I. — The  line  perpendicular  to  the  common 
centre-line  of  two  circles,  and  dividing  the  distance  be- 
tween the  centres  into  parts  such  that  the  difference  of 
their  squares  is  equal  to  the  difference  of  the  squares  on  the 
conterminous  radii,  is  the  radical  axis  of  the  two  circles. 

Cor.  I.  When  two  circles  intersect,  their  radical  axis  is  the 
secant  line  through  the  points  of  intersection. 

Cor.  2.  When  two  circles  touch,  their  radical  axis  is  the 
common  tangent  at  their  point  of  contact. 

Cor.  3.  When  two  circles  are  mutually  exclusive  without 
contact,  their  radical  axis  lies  between  them. 

Cor.  4.  When  two  circles  are  equal  and  concentric,  their 
radical  axis  is  any  line  whatever,  and  when  unequal  and  con- 
centric it  is  the  line  at  00 . 

Def.  2. — When  three  or  more  circles  have  a  common 
radical  axis  they  are  said  to  be  co-axal. 

274°.  If  several  circles  pass  through  the  same  two  points 
they  form  a  co-axal  system. 

For  (273°,  Cor.  i)  the  line  through  the  points  is  the  radical 
axis  of  all  the  circles  taken  in  pairs,  and  is  therefore  the 
common  radical  axis  of  the  system. 

Such  circles  are  called  circles  of  the  common  point  species^ 
contracted  to  f./. -circles. 

Let  a  system  of  ^./.-circles  S,  S^,  S2,  ...,  pass  through  the 
common  points  P  and  O,  and  let  L'L  be  the  right  bisector 
ofPO. 

Then  the  centres  of  all  the  circles  of  the  f./.-system  lie 
on    L'L   and   have    M'M   for  their    common    radical    axis. 


OF  THE   RADICAL  AXIS. 


227 


Hence  from  any  point  C  in   M'M  the  tangents  to  all  the 
circles  are  equal  to  one  another.  (178°) 


Let  CT  be  one  of  these  tangents.  The  circle  Z  with  C  as 
centre  and  CT  as  radius  cuts  all  the  <;./. -circles  orthogonally. 

Similarly,  a  system  of  circles  Z,  Zj,  Zg,  ...  may  be  found 
with  centres  lying  on  M'M  such  that  each  one  of  the  system 
cuts  orthogonally  every  one  of  the  r./. -circles. 

Since  the  centre  of  any  circle  of  this  new  system  is  obtained 
by  drawing  a  tangent  from  any  one  of  the  circles,  as  Sg,  of 
the  <:.^. -species,  to  meet  MM',  it  follows  that  no  circle  of  this 
new  system  can  have  its  centre  lying  between  P  and  Q.  As 
T  approaches  P  the  dependent  circle  Z  contracts  until  it  be- 
comes the  point-circle  P,  when  T  comes  to  coincidence  with  P. 

Hence  P  and  O  are  limiting  forms  of  the  circles  having 
their  centres  on  M'M  and  cutting  the  c.p.-chdes  orthogon- 
ally. The  circles  of  this  second  system  are  consequently 
called  limithig point  circles^  contracted  to  /./.-circles. 

From  the  way  in  which  /./.-circles  are  obtained  we  see  that 
from  any  point  on  L'L  tangents  to  circles  of  the  /./.-system 
are  all  equal,  and  hence  that  L'L  is  the  radical  axis  of  the 
/./.-circles.     Thus  the  two  systems  of  circles  have  their  radi- 


228  SYNTHETIC   GEOMETRY. 

cal  axes  perpendicular,  and  every  circle  of  one  system  cuts 
every  circle  of  the  other  system  orthogonally. 

Hence  P  and  Q  are  inverse  points  with  respect  to  every 
circle  of  the  /./.-system,  and  with  respect  to  any  circle  of 
either  system  all  the  circles  of  the  other  system  invert  into 
themselves. 

If  P  and  Q  approach  L,  the  c.p.-cirdes  separate  and  the 
/./.-circles  approach,  and  when  P  and  Q  coincide  at  L  the 
circles  of  both  species  pass  through  a  common  point,  and  the 
two  radical  axes  become  the  common  tangents  to  the  respec- 
tive systems. 

If  this  change  is  continued  in  the  same  direction,  P  and  Q 
become  imaginary,  and  two  new  limiting  points  appear  on 
the  line  L'L,  so  that  the  former  /./.-circles  become  ^./.-circles, 
and  the  former  ^./.-circles  become  /./.-circles. 

Thus,  in  the  systems  under  consideration,  two  limiting 
points  are  always  real  and  two  imaginary,  except  when  they 
all  become  real  by  becoming  coincident  at  L. 

Cor.  I.  As  the  r./.-circles  and  the  /^.-circles  cut  each  other 
orthogonally,  the  end-points  of  a  diameter  of  any  circle  of  one 
species  are  conjugate  points  with  respect  to  every  circle  of 
the  other  species.  But  a  circle  of  either  species  may  be  found 
to  pass  through  any  given  point  (259°,  Ex.  5).  .'.  the  polars 
of  a  given  point  with  respect  to  all  the  circles  of  either  species 
are  concurrent. 

Cor.  2.  Conversely,  if  the  polars  of  a  variable  point  P  with 
respect  to  three  circles  are  concurrent,  the  locus  of  the  point 
is  a  circle  which  cuts  them  all  orthogonally. 

For  let  Q  be  the  point  of  concurrence.  Then  P  and  O  are 
conjugate  points  with  respect  to  each  of  the  circles.  Hence 
the  circle  on  PQ  as  diameter  cuts  each  of  the  circles  ortho- 
gonally. (270°) 

Cor.  3.  If  a  system  of  circles  is  cut  orthogonally  by  two 
circles,  the  system  is  co-axal. 

For  the  centres  of  the  cutting  circles  must  be  on  the  radical 


OF   THE    RADICAL  AXIS. 


229 


axis  of  all  of  the  other  circles  taken  in  pairs  ;  therefore  they 
have  a  common  radical  axis. 

Cor.  4.  If  two  circles  cut  two  other  circles  orthogonally,  the 
common  centre-line  of  either  pair  is  the  radical  axis  of  the 
other  pair. 

Cor.  5.  Two  /./.-circles  being  given,  a  circle  of  any  required 
magnitude  can  be  found  co-axal  with  them.  But  if  the  circles 
be  of  the  ^./.-species  no  circle  can  be  co-axal  with  them  whose 
diameter  is  less  than  the  distance  between  the  points. 


Exercises. 

1.  Given  two  circles  of  the  /./.-species  to  find  a  circle  with  a 

given  radius  to  be  co-axal  with  them. 

2.  Given  two  circles  of  either  species  to  find  a  circle  to  pass 

through  a  given  point  and  be  co-axal  with  them. 

3.  To  find  a  point  upon  a  given  line  or  circle  such  that  tan- 

gents from  it  to  a  given  circle  may  be  equal  to  its 
distance  from  a  given  point. 

4.  To  find  a  point  whose  distances  from  two  fixed  points  may 

be  equal  to  tangents  from  it  to  two  fixed  circles. 

275°.  Theorem. — The  difference  of  the  squares  on  the  tan- 
gents from  any  point  to  two  circles  is  equal  to  twice  the 
rectangle  on  the  distance  between  the  centres  of  the  circles 
and  the  distance  of  the  point  from  their  radical  axis. 

Let  P  be  the  point,  S  and 
S'  the  circles,  and  LI  their 
radical  axis.     Let  PQ  be  J_ 
to  AB. 
PT^  —  PT'^ 

=  PA2-PB2-(r2-r'^), 
where  r,  r'  are  radii  of  S 
and  S'. 

But,  273°,  Def  I, 

r2-r'2  =  AI=^ 


230  SYNTHETIC   GEOMETRY. 

and  PA'^-PB2  =  AQ2-QB2,  (172°,  i) 

PT2-PT'2  =  AQ2-QB2-(AI2-IB2) 

=  AB(AQ  -  QB)  -  AB(AI  -  IB) 

=  2AB.IQ  =  2AB.PL.  q.e.d. 

This  relation  is  fundamental  in  the  theory  of  the  radical  axis. 

Cor.  I.  When  P  is  on  the  radical  axis  PL=o,  and  the 
tangents  are  equal,  and  when  P  is  not  on  the  radical  axis  the 
tangents  are  not  equal. 

Cor.  2.  The  radical  axis  bisects  all  common  tangents  to  the 
two  circles. 

Cor.  3.  If  P  lies  on  the  circle  S',  PT'  =  o,  and 
PT2  =  2AB.PL, 
.*.  the  square  of  the  tangent  from  any  point  on  one  circle  to 
another  circle  varies  as  the  distance  of  the  point  from  the 
radical  axis  of  the  circles. 

Cor.  4.  If  C  is  the  centre  of  a  circle  S"  passing  through  P 
and  co-axal  with  S',  PT2  =  2AC  .  PL. 

Now,  if  P  could  at  any  time  leave  this  circle  we  would  have 
PT2_pr'2  =  2Ac.pl, 
where  PT"  is  the  tangent  from  P  to  the  circle  S" 

p'r2_p'p2_  p'p"2 

which  is  impossible  finless  PT"=o. 

Hence  the  locus  of  a  point,  which  so  moves  that  the  square 
on  the  tangent  from  it  to  a  given  circle  varies  as  the  distance 
of  the  point  from  a  given  line,  is  a  circle,  and  the  line  is  the 
radical  axis  of  this  circle  and  the  given  circle. 

Cor.  5.  Let  PT'  =  ^  .  PT,  where  ^  is  a  constant.     Then 
PT2-Pr-  =  (i-/&2)px2  =  2Ab.pl, 
p-p2-2AB.  PL 

As  PT2  varies  as  PL,  P  lies  on  a  circle  co-axal  with  S  and  S'. 

.'.  the  locus  of  a  point  from  which  tangents  to  two  given 

circles  are  in  a  constant  ratio  is  a  circle  co-axal  with  the  two. 


OF   THE   RADICAL  AXIS.  23 1 

Exercises. 

1.  In  Cor.  5  what  is  the  position  Y)f  the  locus  for  k  =  o^  /C'=t, 

^  =  >  I,  /&  negative? 

2.  What  is  the  locus  of  a  point  whose  distances  from  two 

fixed  points  are  in  a  constant  ratio  t 

3.  P  and  Q  are  inverse  points  to  the  circle  I,  and  a  line 

through  P  cuts  circle  I  in  A  and  B.  PQ  is  the  internal 
or  external  bisector  of  the  ^AOB,  according  as  P  is 
within  or  without  the  circle. 

4.  P,  Q  are  the  limiting  points  of  the  //^.-circles  S  and  S',  and 

a  tangent  to  S'  at  T  cuts  S       p^^ ^g 

in  A  and  B. 

Then,  considering  P  as  a 
point-circle,  tangents  from 
any  point  on  S  to  P  and  S' 
are  in  a  constant  ratio. 

.-.  AP:AT  =  BP:BT,  and  PT  is  the  external  bi- 
sector of  lAPB.  If  S'  were  enclosed  by  S,  BT  would 
be  an  internal  bisector. 

5.  The  points  of  contact  of  a  common  tangent  to  two  l.p.- 

circles  subtend  a  right  angle  at  either  limiting  point. 

276°.  Theorejn. — The  radical  axes  of  three  circles  taken  in 
pairs  are  concurrent. 

Let  Si,  S2,  S3  denote  the  circles,  and  let  L  be  the  radical 
axis  of  Si  and  S2,  M  of  S2  and  S3,  and  N  of  S3  and  S^ 

L  and  M  meet  at  some  point  O,  from  which  0Ti  =  0T2, 
and  OT2=OT3,  where  OT^  is  the  tangent  from  O  to  S^,  etc., 
0Ti  =  0T3,  and  O  is  on  N, 
.*.     L,  M,  and  N  are  concurrent  at  O. 

Def. — The  point  of  concurrence  of  the  three  radical  axes  of 
three  circles  taken  in  pairs  is  called  the  radical  centre  of  the 
circles. 

Cor.  I.  If  Si,  S2  are  cut  by  a  third  circle  Z,  the  common 
chords  of  Si,  Z  and  S2,  Z  intersect  on  the  radical  axis  of  Si 
and  So. 


232  SYNTHETIC  GEOMETRY. 

Hence  to  find  the  radical  axis  of  two  given  circles  S^  and 
S2,  draw  any  two  circles  Z  -and  Z^  cutting  the  given  circles. 
The  chords  Sj,  Z  and  83,  Z  give  one  point  on  the  radical  axis 
and  the  chords  S^,  Zj  and  S2,  Z^  give  a  second  point. 

Cor.  2.  If  three  circles  intersect  each  other,  their  three 
common  chords  are  concurrent.  (See  249°,  Ex.  i.) 

Cor.  3.  If  a  circle  touches  two  others,  the  tangents  at  the 
points  of  contact  meet  upon  the  radical  axis  of  the  two. 

Cor.  4.  If  a  circle  cuts  three  circles  orthogonally,  its  centre 
is  at  their  radical  centre  and  its  radius  is  the  tangent  from 
the  radical  centre  to  any  one  of  them. 

Cor.  5.  If  in  Cor.  4  the  three  circles  are  co-axal,  any  num- 
ber, of  circles  may  be  found  to  cut  them  orthogonally,  and 
hence  they  have  no  definite  radical  centre,  as  any  point  upon 
the  common  radical  axis  of  the  three  becomes  a  radical 
centre. 

Cor.  6.  If  in  Cor.  4  the  three  circles  mutually  intersect  one 
another,  the  radical  centre  is  within  each  circle  (Cor.  2),  and 
no  tangent  can  be  drawn  from  the  radical  centre  to  any  one 
of  the  circles.  In  this  case  the  circle  which  cuts  them  all  or- 
thogonally has  a  real  centre  but  an  imaginary  radius. 

277°.  Theorem.  — li  ^.ny  three  lines  be  drawn  from  the  ver- 
tices of  a  A  to  the  opposite  sides,  the  polar  centre  of  the  A 
is  the  radical   centre   of  the   circles  having  these  lines  as 

A  R  z B  diameters. 

ABC  is  a  A  and  O  its  ortho- 
centre,  and  AP,  BQ,  CR  are  lines 
from  the  vertices  to  the  opposite 
sides. 
Q      •.•  ^BXA=n, 

o  the    circle    on    AP    as    diameter 

passes  through  X,  and  OX.  OA  is  equal  to  the  square  on  the 
tangent  from  O  to  the  circle  on  AP. 


OF   THE   RADICAL  AXIS.  233 

Similarly  OY.  OB  is  the  square  of  the  tangent  from  O  to 
the  circle  on  BQ  as  diameter,  and  similarly  for  OZ.OC. 
But  as  O  is  the  polar  centre  of  AABC,  (269°) 

OX.  OA  =  OY.  OB  =  OZ.  OC 
.*.  the  tangents  from  O  to  the  three  circles  on  AP,  BQ,  and 
CR  are  equal,  and  O  is  their  radical  centre.  q.e.d. 

Cor.  I.  Let  P,  Q,  R  be  collinear. 

Then  the  polar  centre  of  AABC  is  the  radical  centre  of 
circles  on  AP,  BQ,  and  CR  as  diameters. 

Again,  in  the  AAQR  AP,  QB,  and  RC  are  lines  from  the 
vertices  to  the  opposite  sides. 

.'.  the  polar  centre  of  AAQR  is  the  radical  centre  of  circles 
on  AP,  BQ,  and  CR  as  diameters. 

Similarly  the  polar  centres  of  the  As  BPR  and  CPQ  are 
radical  centres  to  the  same  three  circles. 

But  these  As  have  not  a  common  polar  centre,  as  is  readily 
seen.  Hence  the  same  three  circles  have  four  different 
radical  centres.  And  this  is  possible  only  when  the  circles 
are  co-axal.  (276°,  Cor.  5) 

.*.  the  circles  on  AP,  BQ,  and  CR  are  co-axal. 

.'.  if  any  three  collinear  points  upon  the  sides  of  a  A  be 
joined  with  the  opposite  vertices,  the  circles  on  these  joins  as 
diameters  are  co-axal. 

Cor.  2.  Since  ARPC  is  a  quadrangle  or  tetragram  (247°, 
Def.  2),  and  AP,  BQ,  CR  are  its  three  diagonals, 

.'.  the  circles  on  the  three  diagonals  of  any  quadrangle  are 
co-axal. 

Cor.  3.  The  middle  points  of  AP,  BQ,  and  CR  are  col- 
linear. But  ARPC  is  a  quadrangle  of  which  AP  and  CR  are 
internal  diagonals,  and  BQ  the  external  diagonal. 

.*.  the  middle  points  of  the  diagonals  of  a  complete  quad- 
rangle, or  tetragram,  are  collinear.  (See  248°,  Ex.  2) 

Cor.  4.  The  four  polar  centres  of  the  four  triangles  deter- 
mined by  the  sides  of  a  tetragram  taken  in  threes  are  collinear 


234  SYNTHETIC   GEOMETRY. 

and  lie  upon  the  common  radical  axis  of  the  three  circles 
having  the  diagonals  of  the  tetragram  as  diameters. 

278°.  Theorem. — In  general  a  system  of  co-axal  circles 
inverts  into  a  co-axal  system  of  the  same  species. 

(i.)  Let  the  circles  be  of  the  r./. -species. 

The  common  points  become  two  points  by  inversion,  and 
the  inverses  of  all  the  circles  pass  through  them.  Therefore 
the  inverted  system  is  one  of  ^./.-circles. 

Cor:  I.  The  axis  of  the  system  (LL'  of  Fig.  to  274")  inverts 
into  a  circle  through  the  centre  of  inversion  (261°,  Cor.  i),  and 
as  all  the  inverted  circles  cut  this  orthogonally,  the  axis  of 
the  system  and  the  two  common  points  invert  into  a  circle 
through  the  centre  and  a  pair  of  inverse  points  to  it. 

(258°,  Conv.) 

Cor.  2.  If  one  of  the  common  points  be  taken  as  the  centre 
of  inversion,  its  inverse  is  at  00 . 

The  axis  of  the  system  then  inverts  into  a  circle  through 
the  centre  of  inversion,  and  having  the  inverse  of  the  other 
common  point  as  its  centre,  and  all  the  circles  of  the  system 
invert  into  centre-lines  to  this  circle. 

(2.)  Let  the  circles  be  of  the  /./.-species. 

Let  the  circles  S  and  S'  pass  through  the  limiting  points, 
and  be  thus  ^./.-circles. 

Generally  S  and  S'  invert  into  circles  which  cut  the  in- 
verses of  all  the  other  circles  orthogonally.  (264°) 

.'.  the  intersections  of  the  inverses  of  S  and  S'  are  limiting 
points,  and  the  inverted  system  is  of  the  /./.-species. 

Cor.  3.  The  axis  of  the  system  (MM'  of  Fig.  to  274°)  be- 
comes a  circle  through  the  centre  and  passing  through  the 
limiting  points  of  the  inverted  system,  thus  becoming  one  of 
the  r./. -circles  of  the  system. 

Cor.  4.  If  one  of  the  limiting  points  be  made  the  centre  of 


OF   THE  RADICAL  AXIS. 


235 


inversion,  the  circles  S  and  S'  become  centre-lines,  and  the 
/./.-circles  become  concentric  circles. 

Hence  concentric  circles  are  co-axal,  their   radical  axis 
being  at  00 . 

Exercises. 

1.  What  does  the  radical  axis  of  (i,  278^^)  become? 

2.  What  does  the  radical  axis  of  (2,  278°)  become  ? 

3.  How  would  you  invert  a  system  of  concentric  circles  into 

a  common  system  of  /./.-circles  ? 

4.  How  would  you  invert  a  pencil  of  rays  into  a  system  of 

^./.-circles. 

5.  The  circles  of  277°  are  common  point  circles. 


279".   Theorem. — Any  two    circles    can  be  inverted  into 
equal  circles. 

Let  S,  S'  be  the  circles 
having  radii  r  and  r',  and 
let  C,  C  be  the  equal 
circles  into  which  S  and 
S'  are  to  be  inverted  ;  and 
let  the  common  radius  be  p. 

But,  since  P  and  Q  and  also  P'  and  Q'  are  inverse  points, 
OP.OO  =  OP'.OQ', 
OQ2     r  ,     , 

and  (275°,  Cor.  5)  O  lies  on  a  circle  co-axal  with  S  and  S'. 
And  with  any  point  on  this  circle  as  a  centre  of  inversion  S 
and  S'  invert  into  equal  circles. 

Cor.  I.  Any  three  non-co-axal  circles  can  be  inverted  into 
equal  circles. 


\l6 


SYNTHETIC   GEOMETRY. 


For,  let  the  circles  be  S,  S',  S",  and  let  Z  denote  the  locus 
of  O  for  which  S  and  S'  invert  into  equal  circles,  and  Z'  the 
locus  of  O  for  which  S  and  S"  invert  into  equal  circles.    Then 


Z'  is  co-axal  with  S  and  S".  And,  as  S,  S',  and  S"  are  not 
co-axal,  Z  and  Z'  intersect  in  two  points,  with  either  of  which 
as  centre  of  inversion  the  three  given  circles  can  be  inverted 
into  equal  circles. 

Cor.  2.  If  S,  S',  and  S"  be  /./.-circles,  Z  and  Z'  being 
co-axal  with  them  cannot  intersect,  and  no  centre  exists  with 
which  the  three  given  circles  can  be  inverted  into  equal  circles. 

But  if  S,  S'  and  S"  be  ^/.-circles,  Z  and  Z'  intersect  in  the 
common  points,  and  the  given  circles  invert  into  centre-lines 
of  the  circle  of  inversion,  and  having  each  an  infinite  radius 
these  circles  may  be  considered  as  being  equal.   (278°,  Cor.  2) 

Cor.  3.  In  general  a  circle  can  readily  be  found  to  touch 
three  equal  circles.  Hence  by  inverting  a  system  of  three 
circles  into  equal  circles,  drawing  a  circle  to  touch  the  three, 
and  then  re-inverting  we  obtain  a  circle  which  touches  three 
given  circles. 

If  the  three  circles  are  co-axal,  no  circle  can  be  found  to 
touch  the  three. 

280°.  Let  the  circles  S  and  S',  with  centres  A  and  B  and 

radii  r  and  r,  be  cut 
by  the  circle  Z  with 
centre  at  O  and  radius 
OP  =  R.  Let  NLbe 
the  radical  axis  of  S 
and  S'. 

Since  AP  is  J.  to 
the  tangent  at  P  to 
the  circle  S,  and  OP 
is  JL  to  the  tangent 
at  P  to  the  circle  Z, 

the  ^APO  =  ^  is  the  angle  of  intersection  of  the  circles  S 


OF   THE   RADICAL  AXIS.  53/ 

and  Z  (115°,  Def.  i).     Similarly  BQO-0  is  the  angle  of 
intersection  of  the  circles  S'  and  Z.     Now 

PP'  =  2rcos^  =  R-OP', 
and  QQ'  =  2r'cos0  =  R-OQ', 

OP' -  OQ' =  2(r'cos  0  -  r  cos  ^). 
But  R.OP'-R.OQ'  =  OT2-Or2    (where   OT  is  the 

tangent  from  O  to  S,  etc.)  =2AB .  OL,  (275°) 

R  =  _ ^ .OL. 

rcos<i>-rcosd 

Cor.  I.  When  6  and  <p  are  constant,  R  varies  as  OL. 

.*.  a  variable  circle  which  cuts  two  circles  at  constant 
angles  has  its  radius  varying  as  the  distance  of  its  centre  from 
the  radical  axis  of  the  circles. 

Cor.  2.  Under  the  conditions  of  Cor.  i  ON  varies  as  OL, 

and  .*.  --^,  is  constant. 
ON 

.•.a  variable  circle  which  cuts  two  circles  at  a  constant 

angle  cuts  their  radical  axis  at  a  constant  angle. 

Cor.  3.  When  0L  =  o,  /coscp^rcosd, 
and  r  :  r'  =  cos0  :  cos^. 

.'.  a  circle  with  its  centre  on  the  radical  axis  of  two  other 
circles  cuts  them  at  angles  whose  cosines  are  inversely  as  the 
radii  of  the  circles. 

Cor.  4.  If  circle  Z  touches  S  and  S',  d  and  <p  are  both  zero 

ot  both  equal  to  tt,  or  one  is  zero  and  the  other  is  tt. 

AB 
.'.  when  Z  touches  S  and  S'  R=  .    ,       .  OL,  where  the 

variation  in  sign  gives  the  four  possible  varieties  of  contact. 

Cor.  5.  When  d  =  <f)  =  2'  Z  cuts  S  and  S'  orthogonally,  and 
OL=o,  and  the  centre  of  the  cutting  circle  is  on  the  radical 
axis  of  the  two. 


238 


SYNTHETIC   GEOMETRY. 


SECTION   VII. 

CENTRE    AND    AXES    OF    SIMILITUDE    OR 
PERSPECTIVE. 

The  relations  of  two  triangles  in  perspective  have  been 
given  in  Art.  254°.  We  here  propose  to  extend  these  rela- 
tions to  the  polygon  and  the  circle. 


Let  O,  any  point,  be  connected  Avith  the  vertices  A, 
B,   C,  ...  of  a  polygon,  and  on 
OA,  OB,  OC,  ...  let  points  a,  b, 
c,  ...  be  taken  so  that 
OA  :  0^  =  0B  :  Ol?=OC  :  Or... 
and 

OA  :  0^'  =  0B  :  0^'  =  0C  :  O^:'... 
Then,  since  OAB  is  a  A  and  ab 
is  so  drawn  as  to  divide  the  sides 
proportionally  in  the  same  order, 
.-.  ab  is  II  to  AB.       (202°,  Conv.) 
Similarly, 
be  is  II  to  BC,      cd  to  CD,  etc., 
b'c'  is  II  to  BC,    (fd'  to  CD,  etc., 
AO AB  ^  /\Oab  «  /Spa'b', 
AOl^C^ /\Obc^ AOb'c',  ... 
.*.  the  polygons  ABC...,  abc...,  and  a'b'c'...   are  all  similar 
and  have  their  homologous  sides  parallel. 

Def.—The.  polygons  ABCD...  and  abed...  are  said  to  be 
similarly  placed,  and  O  is  their  ^aY^r;/^?/ centre  of  similitude; 
while  the  polygons  ABCD...  and  a'b'e'd'...  are  oppositely 
placed,  and  O  is  their  internal  centre  of  similitude. 

Hence,  when  the  lines  joining  any  point  to  the  vertices  of 
a  polygon  are  all  divided  in  the  same  manner  and  in  the 
same  order,  the  points  of  division  are  the  vertices  of  a  second 


similarly, 
and 


CENTRE  OF  SIMILITUDE  OR  PERSPECTIVE.     239 

polygon   similar  to   the   original,   and    so    placed  that   the 
homologous  sides  of  the  two  polygons  are  parallel. 


282°.  When  two  similar  polygons  are  so  placed  as  to  have 
their  homologous  sides  parallel,  they  are  in  perspective,  and 
the  joins  of  corresponding  vertices  concur  at  a  centre  of 
similitude. 

Let  ABCD...,  abed...  be  the  polygons. 

Since  they  are  similar,  AB  :  ab=^^C  :  bc=CD  :  cd...  (207"), 
and  by  hypothesis  AB  is  ||  to  ab,  BC  to  bc^  etc. 

Let  ha  and  ^b  meet  at  some  point  O. 

Then  GAB  is  a  A  and  ab\s\\  to  AB. 
OB_AB_BC_   .^ 
Wb      ab       be  ' 

.".  C<;  passes  through  O,  and  similarly  D^^  passes  through  O, 
etc. 

By  writing  alb'd ...  for  abe...  the  theorem  is  proved  for  the 
polygon  a'b'e'd',  which  is  oppositely  placed  to  ABCD... 

Cor.  I.  If  Aa  and  V>b  meet  at  00,  <?^  =  AB,  and  hence 
^(r=BC,  etc.,  and  the  polygons  are  congruent. 

Cor.  2.  The  joins  of  any  two  corresponding  vertices  as  A, 
C  ;  a,  c  ',  a\  e'  are  evidently  homologous  lines  in  the  polygons 
and  are  parallel. 

Similarly  any  line  through  the  centre  O,  as  XxOx'  is 
homologous  for  the  polygons  and  divides  them  similarly. 

283°.  Let  the  polygon  ABCD...  have  its  sides  indefinitely 
increased  in  number  and  diminished  in  length.  Its  limiting 
form  (148°)  is  some  curve  upon  which  its  vertices  lie.  A 
similar  curve  is  the  limiting  form  of  the  polygons  abed...  as 
also  oi  a'b'c'd'...,  since  every  corresponding  pair  of  limiting  or 
vanishing  elements  are  similar. 

Hence,  if  two  points  en  a  variable  radius  vector  have  the 
ratio  of  their  distances  from  the  pole  constant,  the  loci  of  the 


240  •  SYNTHETIC   GEOMETRY. 

points  are  similar  curves  in  perspective,  and  having  the  pole 
as  a  centre  of  perspective  or  simihtude. 

Cor.  I.  In  the  limiting  form  of  the  polygons,  the  line  BC 
becomes  a  tangent  at  B,  and  the  line  be  becomes  a  tangent  at 
b.     And  similarly  for  the  line  be'. 

.'.  the  tangents  at  homologous  points  on  any  two  curves  in 
perspective  are  parallel. 

284°.  Since  abed. . .  and  a'b'e'd'. . .  are  both  in  perspective  with 
ABCD...  and  similar  to  it,  we  see  that  two  similar  polygons 
may  be  placed  in  two  different  relative  positions  so  as  to  be 
in  perspective,  that  is,  they  may  be  similarly  placed  or  oppo- 
sitely placed. 

In  a  reguLir  polygon  of  an  even  number  of  sides  no  dis- 
tinction can  be  made  between  these  two  positions  ;  or,  two 
similar  regular  polygons  are  both  similarly  and  oppositely 
placed  at  the  same  time  when  so  placed  as  to  be  in  per- 
spective. 

Hence  two  regular  polygons  of  an  even  number  of  sides 
and  of  the  same  species,  when  so  placed  as  to  have  their  sides 
respectively  parallel,  have  two  centres  of  perspective,  one  due 
to  the  polygons  being  similarly  placed,  the  external  centre  ; 
and  the  other  due  to  the  polygons  being  oppositely  placed, 
the  internal  centre. 

Cor.  Since  the  limiting  form  of  a  regular  polygon  is  a  circle 
(148°),  two  circles  are  always  similarly  and  oppositely  placed 
at  the  same  time,  and  accordingly  have  always  two  centres  of 
perspective  or  similitude. 

285°.  Let  S  and  S'  be  two  circles  with  centres  C,  C  and 
radii  r,  r'  respectively,  and  let  O  and  O'  be  their  centres  of 
perspective  or  similitude. 

Let  a  secant  line  through  O  cut  S  in  X  and  Y,  and  S'  in  X' 
and  Y'. 


CENTRE  OF  SIMILITUDE  OR  PERSPECTIVE.     241 
Then  O  is  the  centre  of  similitude  due  to  considering  the 


circles  S  and  S'  as  being 
similarly  placed. 

Hence  X  and  X',  as  also 
Y  and  Y',  are  homologous 
points,  and  (283°,  Cor.  i) 
the  tangents  at  X  and  X' 
are  parallel.  So  also  the 
tangents  at  Y  and  Y'  are 
parallel. 

Again  O'  is  the  centre 
of  similitude  due  to  con- 
sidering the  circles  as 
being  oppositely  placed, 
and  for  this  centre  Z  and  Y'  as  also  U  and  U'  are  homologous 
points  ;  and  tangents  at  Y'  and  Z  are  parallel,  and  so  also 
are  tangents  at  U  and  U'. 

Hence  YZ  is  a  diameter  of  the  circle  S  and  is  parallel  to 
Y'Z'  a  diameter  of  the  circle  S'. 

Hence  to  find  the  centres  of  similitude  of  two  given 
circles  : — Draw  parallel  diameters,  one  to  each  circle,  and 
connect  their  end-points  directly  and  transversely.  The 
direct  connector  cuts  the  common  centre-line  in  the  external 

Q 


242  SYNTHETIC   GEOMETRY. 

centre  of  similitude,  and  the  transverse  connector  cuts  it  in 
the  internal  centre  of  simihtude. 

286°.  Since  OX  :  OX'  =  OY  :  OY',  if  X  and  Y  become  coin- 
cident, X'  and  Y'  become  coincident  also. 

.'.  a  line  through  O  tangent  to  one  of  the  circles  is  tangent 
to  the  other  also,  or  O  is  the  point  where  a  common  tangent 
cuts  the  common  centre-line.     A  similar  remark  applies  to  O'. 

When  the  circles  exclude  one  another  the  centres  of 
similitude  are  the  intersections  of  common  tangents  of  the 
same  name,  direct  and  transverse. 

When  one  circle  lies  within  the  other  (2nd  Fig.)  the  com- 
mon tangents  are  imaginary,  although  O  and  O'  their  points 
of  intersection  are  real. 

287°.  Since  AOCY^AOC'Y',      .-.    OC:OC  =  r:?'', 
and  since  AO'CZ^  AO'C'Y',     .-.  O'C  :  0'C'  =  r  :  r\ 

.'.  the  centres  of  similitude  of  two  circles  are  the  points 
which  divide,  externally  and  internally,  the  join  of  the  centres 
of  the  circles  into  parts  which  are  as  the  conterminous  radii. 

The  preceding  relations  give 

OC  =  -/!-.  CC,  and  0'C=-;^     .  CC. 
r  -J'  r  -\-r 

OC  is  <  r  according  as  CC  is  <  r'  —  r, 

and       O'C  is  ^  r  according  as  CC  is  %  r'  +  r. 

Hence 

1.  O  lies  within  the  circle  S  when  the  distance  between  the 
centres  is  less  than  the  difference  of  the  radii,  and  O'  lies 
within  the  circle  S  when  the  difference  between  the  centres  is 
less  than  the  sum  of  the  radii. 

2.  When  the  circles  exclude  each  other  without  contact  both 
centres  of  similitude  lie  without  both  circles. 

3.  When  the  circles  touch  externally,  the  point  of  contact 
is  the  internal  centre  of  similitude. 

4.  When  one  circle  touches  the  other  internally,  the  point 
of  contact  is  the  external  centre  of  similitude. 


CENTRE  OF  SIMILITUDE  OR  PERSPECTIVE.     243 

5.  When  the  circles  are  concentric,  the  centres  of  similitude 
coincide  with  the  common  centre  of  the  circles,  unless  the 
circles  are  also  equal,  when  one  centre  of  similitude  becomes 
any  point  whatever. 

6.  If  one  of  the  circles  becomes  a  point,  both  centres  of 
similitude  coincide  with  the  point. 

288°.  Def. — The  circle  having  the  centres  of  similitude  of 
two  given  circles  as  end-points  of  a  diameter  is  called  the 
circle  of  similitude  of  the  given  circles. 

The  contraction  0  of  s.  will  be  used  for  circle  of  similitude. 

Cor.  I.  Let  S,  S'  be  two  circles  and  Z  their  0  of  s. 

Since  O  and  O'  are 
two  points  from  which 
tangents  to  circles  S  and 
S'  are  in  the  constant 
ratio  of  r  to  r',  the  circle  o" 
Z  is  co-axal  with  S  and 
S'  (275°,  Cor.  5).  Hence 
any  two  circles  and  their  Q)  of  s.  are  co-axal. 

Cor.  2.  From  any  point  P  on  circle  Z, 
PT  :TC  =  PT':T'C, 
and.-.  ^TPC  =  ^T'PC'. 

Hence,  at  any  point  on  the  Q)  of  s.  oi  two  circles,  the  two 
circles  subtend  equal  angles. 

Cor.  3.      OC  =  CC.  -,^.  and  O'C  =  CC.  -^.  ^287°) 

^  r'-r  r'  +  r  '     '  ^ 

whence       00'=  CC. 


The  Q)  of  s.\s  a  line,  the  radical  axis,  when  the  given 
circles  are  equal  {r=r'). 

The  (z)  of  s.  becomes  a  point  when  one  of  the  two  given 
circles  becomes  a  point  (r  or  r'  =  o). 

The  0  ofs.  is  a  point  when  the  given  circles  are  con- 
centric (CC'=o). 


244  SYNTHETIC   GEOMETRY. 

289°.  Def. — With  reference  to  the  centre  O  (Fig.  of  285°), 
X  and  y,  as  also  X'  and  Y,  are  called  antihomologoiis  points. 
Similarly  with  respect  to  the  centre  O',  U'  and  Z,  as  also  U 
and  Y',  are  antihomologous  points. 

Let  tangents  at  X  and  Y'  meet  at  L.  Then,  since  CX  is  || 
to  CX',  ^CXY=^C'X'Y'  =  z.C'Y'X'.  But  ^LXY  is  comp.  of 
^CXY  and  Z.LY'X'  is  comp.  of  Z.C'Y'X'. 

ALXY' is  isosceles,  and  LX  =  LY'. 
L  is  on  the  radical  axis  of  S  and  S'. 

Similarly  it  may  be  proved  that  pairs  of  tangents  at  Y  and 
X',  at  U  and  Y',  and  at  U'  and  Z,  meet  on  the  radical  axis  of 
S  and  S',  and  the  tangent  at  U  passes  through  L. 

.'.  tangents  at  a  pair  of  antihomologous  points  meet  on  the 
radical  axis.- 

Cor.  I.  The  join  of  the  points  of  contact  of  two  equal 
tangents  to  two  circles  passes  through  a  centre  of  similitude 
of  the  two  circles. 

Cor.  2.  When  a  circle  cuts  two  circles  orthogonally,  the 
joins  of  the  points  of  intersection  taken  in  pairs  of  one  from 
each  circle  pass  through  the  centres  of  similitude  of  the  two 
circles. 

290°.  Since  OX:OX'  =  r:r', 

OX.OY':OX'.  OY'  =  r:r'. 
But  OX'.  OY'  =  the  square  of  the  tangent  from  O  to  the  circle 
S'  and  is  therefore  constant. 

OX  .  0Y'  =  -, .  0T'2  =  a  constant. 
r 

.'.  X  and  Y'  are  inverse  points  with  respect  to  a  circle 
whose  centre  is  at  O  and  whose  radius  is  OT'     /^,. 

B^/. — This  circle  is  called  the  circle  of  antisimilihide^  and 
will  be  contracted  to  0  ofans. 

Evidently  the  circles  S  and  S'  are  inverse  to  one  another 
with  respect  to  their  0  ofans. 


CENTRE  OF  SIMILITUDE  OR  PERSPECTIVE.     245 

For  the  centre  0'  the  product  OU  .  OY'  is  negative,  and  the 
0  of  ans.  corresponding  to  this  centre  is  imaginary. 

Cor.  I.  Denoting  the  distance  CC  by  d^  and  the  difference 
between  the  radii  {r'  —  r)  by  5,  we  have 

R'  =  ^^'-— 52-' 

where  R  =  the  radius  of  the  0  of  ans.     Hence 

1.  When  either  circle  becomes  a  point  their  0  of  ans. 

becomes  a  point. 

2.  When  the  circles  S  and  S'  are  equal,  the  0  of  ans.  be- 

comes the  radical  axis  of  the  two  circles. 

3.  When  one  circle  touches  the  other  internally  the  0  of 

ans.  becomes  a  point-circle.     (d=d.) 

4.  When  one  circle  includes  the  other  without  contact  the 

0  of  ans.  is  imaginary.     (d<d.) 

Cor.  2.  Two  circles  and  their  circle  of  antisimilitude  are 
co-axal.  (263°) 

Cor.  3.  If  two  circles  be  inverted  with  respect  to  their 
circle  of  antisimilitude,  they  exchange  places,  and  their  radi- 
cal axis  being  a  line  circle  co-axal  with  the  two  circles 
becomes  a  circle  through  O  co-axal  with  the  two. 

The  only  circle  satisfying  this  condition  is  the  circle  of 
similitude  of  the  two  circles.  Therefore  the  radical  axis 
inverts  into  the  circle  of  similitude,  and  the  circle  of  simili- 
tude into  the  radical  axis. 

Hence  every  line  through  O  cuts  the  radical  axis  and  the 
circle  of  similitude  of  two  circles  at  the  same  angle. 

291°.  D^f — When  a  circle  touches  two  others  so  as  to 
exclude  both  or  to  include  both,  it  is  said  to  ifouc/i  them 
similarly^  or  to  have  contacts  of  like  kind  with  the  two. 
When  it  includes  the  one  and  excludes  the  other,  it  is  said  to 
touch  them  dissimilarly^  or  to  have  contacts  of  unlike  kinds 
with  the  two. 


246 


SYNTHETIC   GEOMETRY. 


292°.  Theorem. — When  a  circle  touches  two  other  circles, 
its  chord  of  contact  passes  through  their  external  centre  of 
similitude  when  the  contacts  are  of  like  kind,  and  through 
their  internal  centre  of  similitude  when  the  contacts  are  of 
unlike  kinds. 


Proof. — Let  circle  Z  touch  circles  S  and  S'  at  Y  and  X'. 
Then  CYD  and  C'X'D  are  lines.  (113°,  Cor.  i) 

Let  XYX'Y'  be  the  secant  through  Y  and  X'.     Then 
^CXY  =  aCYX  =  ^DYX'  =  ^DX'Y  =  ^CX'Y'. 
.*.  CX  and  C'X'  are  parallel,  and  X'X  passes  through  the 
external  centre  of  similitude  O.  (285") 

Similarly,  if  Z'  includes  both  S  and  S',  it  may  be  proved 
that  its  chord  of  contact  passes  through  O. 

Again,  let  the  circle  W,  with  centre  E,  touch  S'  at  Y'  and 
S  at  U  so  as  to  include  S'  and  exclude  S,  and  let  UY'  be  the 
chord  of  contact.     Then 

^CVU  =  i.CUV  =  ^EUY'  =  _EY'U, 
.*.  EY'  and  CV  are  parallel  and  VY'  connects  them  trans- 
versely ;  .*.  VY'  passes  through  O'.  q.e.d. 


Cor.  I.  Every  circle  which  touches  S  and  S'  similarly  is 
cut  orthogonally  by  the  external  circle  of  antisimilitude  of  S 
and  S'. 


CENTRE  OF  SIMILITUDE  OR  PERSPECTIVE.     247 

Cor.  2.  If  two  circles  touch  S  and  S'  externally  their  points 
of  contact  are  concyclic.  (n6°,  Ex.  2) 

But  the  points  of  contact  of  either  circle  with  S  and  S'  are 
antihomologous  points  to  the  centre  O. 

.'.  if  a  circle  cuts  two  others  in  a  pair  of  antihomologous 
points  it  cuts  them  in  a  second  pair  of  antihomologous  points. 

Cor.  3.  If  two  circles  touch  two  other  circles  similarly,  the 
radical  axis  of  either  pair  passes  through  a  centre  of  simili- 
tude of  the  other  pair. 

For,  if  Z  and  Z'  be  two  circles  touching  S  and  S'  externally, 
the  external  circle  of  antisimilitude  of  S  and  S'  cuts  Z  and  Z' 
orthogonally  (Cor.  i)  and  therefore  has  its  centre  on  the 
radical  axis  of  Z  and  Z'. 

Cor.  4.  If  any  number  of  circles  touch  S  and  S'  similarly, 
they  are  all  cut  orthogonally  by  the  external  circle  of  anti- 
similitude  of  S  and  S',  and  all  their  chords  of  contact  and  all 
their  chords  of  intersection  with  one  another  are  concurrent 
at  the  external  centre  of  antisimilitude  of  S  and  S'. 


293°.  71ieorein.—\{  the  circle  Z  touches  the  circles  S  and 
S',  the  chord  of  contact  of  Z  and  the  radical  axis  of  S  and  S' 
are  conjugate  lines  with  respect  to  the  circle  Z. 

Proof. — Let  Z  touch  S  and  S'  in  Y  and  X'  respectively. 
The  tangents  at  Y  and  X'  meet  at  a  point  P  on  the  radical 
axis  of  S  and  S'.  (178  ) 

But  P  is  the  pole  of  the  chord  of  contact  YX'. 

.*.  the  radical  axis  passes  through  the  pole  of  the  chord  of 
contact,  and  reciprocally  the  chord  of  contact  passes  through 
the  pole  of  the  radical  axis  (267°,  Def )  and  the  lines  are 
conjugate.  q.c.d. 


248 


SYNTHETIC   GEOMETRY. 


AXES    OF    SIMILITUDE. 


294°.  Let  Sj,  S2,  S3  denote  three  circles  having  their  centres 
A5  B,  C  and  radii  r^,  ?^2y  ^z^  ^"<^  let  X,  X',  Y,  Y',  Z,  Z'  be  their 
six  centres  of  similitude. 

Now   X,   Y,   Z   are 

three    points    on    the 

sides   of  the  AABC, 

BX^r, 

CX     r.l 


and 


_'3 


r^y 


CY 
AY 
AZ 
BZ 

Vex/     ' 

and  X,  Y,  Z  are  col- 
linear. 

Similarly  it  is  proved  that  the  triads  of  points  XY'Z',  YZ'X', 
ZX'Y'  are  collinear. 

Def. — These  lines  of  coUinearity  of  the  centres  of  similitude 
of  the  three  circles  taken  in  pairs  are  the  axes  of  similitude 
of  the  circles.  The  line  XYZ  is  the  external  axis,  as  being 
external  to  all  the  circles,  and  the  other  three,  passing  be- 
tween the  circles,  are  internal  axes. 

Cor.  I.  If  an  axis  of  similitude  touches  any  one  of  the 
circles  it  touches  all  three  of  them.  (286°) 

Cor.  2.  If  an  axis  of  similitude  cuts  any  one  of  the  circles 
it  cuts  all  three  at  the  same  angle,  and  the  intercepted  chords 
are  proportional  to  the  corresponding  radii. 

Cor.  3.  Since  XYX'Y'  is  a  quadrangle  whereof  XX',  YY', 
and  ZZ'  are  the  three  diagonals,  the  circles  on  XX',  YY',  and 
ZZ'  as  diameters  are  co-axal.  (277°,  Cor.  2) 

.'.  the  circles  of  similitude  of  three  circles  taken  in  pairs 
are  co-axal. 


AXES  OF  SIMILITUDE  OR  PERSPECTIVE.      249 

Cor.  4.  Since  the  three  circles  of  similitude  are  of  the  c.p.- 
species,  two  points  may  be  found  from  which  any  three  circles 
subtend  equal  angles.  These  are  the  common  points  to  the 
three  circles  of  similitude.  (288°,  Cor.  2) 

Cor.  5.  The  groups  of  circles  on  the  following  triads  of 
segments  as  diameters  are  severally  co-axal, 
AX,  BY,  CZ  ;  AX,  YZ',  Y'Z  ;  BY,  Z'X,  ZX';  CZ,  XY',  X'Y. 

295°.  Any  two  circles  Z  and  Z',  which  touch  three  circles 
Si,  S2,  S3  similarly,  cut  their  circles  of  antisimilitude  ortho- 
gonally (292°,  Cor.  i),  and  therefore  have  their  centres  at  the 
radical  centre  of  the  three  circles  of  antisimilitude. 

(276°,  Cor.  4) 

But  Z  and  Z'  have  not  necessarily  the  same  centre. 
.*.  the  three  circles  of  antisimilitude  of  the  circles  S^,  S2,  and 
S3  are  co-axal,  and  their  common  radical  axis  passes  through 
the  centres  of  Z  and  Z'. 

296°.  Theorem. — If  two  circles  touch  three  circles  similarly, 
the  radical  axis  of  the  two  is  an  axis  of  similitude  of  the 
three  ;  and  the  radical  centre  of  the  three  is  a  centre  of 
similitude  of  the  two. 

Proof. — The  circles  S  and  S' 
touch  the  three  circles  A,  B, 
and  C  similarly. 

I.  Since  S   and   S'  touch   A 

and  B  similarly,  the  radical  axis 

of  S  and  S'  passes  through  a 

centre  of  similitude  of  A  and  B. 

(292°,  Cor.  3) 

Also,  the   radical   axis   of   S 
•  and  S'  passes  through  a  centre  of  similitude  of  B  and  C,  and 
through  a  centre  of  similitude  of  C  and  A. 

,*.  the  radical  axis  of  S  and  S'  is  an  axis  of  similitude  of 
the  three  circles  A,  B,  and  C. 


250  SYNTHETIC   GEOMETRY. 

2.  Again,  since  A  and  B  touch  S  and  S',  the  radical  axis  of 
A  and  B  passes  through  a  centre  of  simiHtude  of  S  and  S'. 

For  similar  reasons,  and  because  A,  B,  and  C  touch  S  and 
S'  similarly,  the  radical  axes  of  B  and  C,  and  of  C  and  A, 
pass  through  the  same  centre  of  similitude  of  S  and  S'.  But 
these  three  radical  axes  meet  at  the  radical  centre  of  A,  B, 
and  C. 

.',  the  radical  centre  of  A,  B,  and  C  is  a  centre  of  simili- 
tude of  S  and  S'.  q.e.d. 

297°.  Problem. — To  construct  a  circle  which  shall  touch 
three  given  circles. 

In  the  figure  of  296°,  let  A,  B,  and  C  be  the  three  given 
circles,  and  let  S  and  S'  be  two  circles  which  are  solutions 
of  the  problem. 

Let  L  denote  one  of  the  axes  of  similitude  of  A,  B,  and  C, 
and  let  O  be  their  radical  centre.  These  are  given  when  the 
circles  A,  B,  and  C  are  given. 

Now  L  is  the  radical  axis  of  S  and  S'  (296°,  i),  and  O  is 
one  of  their  centres  of  similitude. 

But  as  A  touches  S  and  S'  the  chord  of  contact  of  A  passes 
through  the  pole  of  L  with  respect  to  A  (293°).  Similarly  the 
chords  of  contact  of  B  and  C  pass  through  the  poles  of  L 
with  respect  to  B  and  C  respectively.  And  these  chords  are 
concurrent  at  O.  (292°) 

Hence  the  following  construction  : — 

Find  O  the  radical  centre  and  L  an  axis  of  simiHtude  of  A, 
B,  and  C.  Take  the  poles  of  L  with  respect  to  each  of  these 
circles,  and  let  them  be  the  points  p,  q,  r  respectively. 

Then  Op,  Og,  Or  are  the  chords  of  contact  for  the  three 
given  circles,  and  three  points  being  thus  found  for  each  of 
two  touching  circles,  S  and  S',  these  circles  are  determined. 

(This  elegant  solution  of  a  famous  problem  is  due  to  M. 
Gergonne.) 

Cor.  As  each  axis  of  similitude  gives  different  poles  with 
respect  to  A,  B,  and  C,  while  there  is  but  one  radical  centre 


AXES  OF  SIMILITUDE  OR  PERSPECTIVE.        25  I 

O,  in  general  each  axis  of  similitude  determines  two  touching 
circles ;  and  as  there  are  four  axes  of  similitude  there  are 
eight  circles,  in  pairs  of  twos,  which  touch  three  given  circles. 
Putting  i  and  e  for  internal  and  external  contact  with  the 
touching  circle,  we  may  classify  the  eight  circles  as  follows  : 

(See  294°) 


Axes  of  Similituul. 

A 

B 

c 

X     Y 

Z 

i 
e 

i 
e 

:]'- 

X     Y' 

Z' 

e 
i 

i 

e 

:]^>- 

X'     Y 

Z' 

i 

e 

e 
i 

:]^- 

X'     Y' 

Z 

i 
e 

i 
e 

:.}.pr. 

PART  V. 


ON  HARMONIC  AND  ANHARMONIC  RATIOS— 
HOMOGRAPHY.  INVOLUTION,  ETC. 


SECTION    I. 

GENERAL    CONSIDERATIONS     IN     REGARD    TO 
HARMONIC  AND  ANHARMONIC  DIVISION. 

298°.  Let  C  be  a  point  dividing  a  segment  AB.  The  posi- 
tion of  C  in  relation  to  A  and  B  is  determined  by  the  ratio 

, ^         ,  AC  :  BC.     For,  if  we  know  this  ratio,  we 

^  0        D         B  know  completely  the  position  of  C  with 

respect  to  A  and  B.  If  this  ratio  is  negative,  C  lies  between 
A  and  B  ;  if  positive,  C  does  not  lie  between  A  and  B.  If 
AC:BC=  — I,  C  is  the  internal  bisector  of  AB  ;  and  if 
AC  :  BC=  +  1,  C  is  the  external  bisector  of  AB,  i.e.,  a  point 
at  CO  in  the  direction  AB  or  BA. 

Let  D  be  a  second  point  dividing  AB.  The  position  of  D 
is  known  when  the  ratio  AD  :  BD  is  known. 

Be/. — If  we  denote  the  ratio  AC  :  BC  by  w,  and  the  ratio 
AD  :  BD  by  ;/,  the  two  ratios  7/1 :  n  and  n  :  m,  which  are 
reciprocals  of  one  another,  are  called  the  two  anhannonic 
ratios  of  the  division  of  the  segment  AB  by  the  points  C  and 
D,  or  the  harmonoids  of  the  range  A,  B,  C,  D. 
252 


HARMONIC  AND  ANHARMONIC  DIVISION.     253 

Either  of  the  two  anharmonic  ratios  expresses  a  relation 
between  the  parts  into  which  the  segment  AB  is  divided  by 
the  points  C  and  D. 

Evidently  the  two  anharmonic  ratios  have  the  same  sign, 
and  when  one  of  them  is  zero  the  other  is  infinite,  and  vice 
versa. 

These  ratios  may  be  written  : — 

AC  .  AD    AC.BD   AC  .  ED 
BC  '  BD    BC.AD   AD  .  BC* 

AD  .  AC   AD^C   AI:K^ 
BD  '  BC    BD.AC   AC.BD* 

The  last  form  is  to  be  preferred,  other  things  being  convenient, 
on  account  of  its  symmetry  with  respect  to  A  and  B,  the 
end-points  of  the  divided  segment. 


299°.  The  following  results  readily  follow. 

T    ,  AC.BD,       ,       T^t,      AC       1  AD  ,         ,.1 

1.  Let  ^^^-^S7^  be  +.     Then  --  and  — -^  have  like  signs, 

AD  .  riC  x>v^  -dU 

and  therefore  C  and  D  both  divide  AB  internally  or  both 
externally.  (298") 

In  this  case  the  order  of  the  points  must  be  some  one  of 
the  following  set,  where  AB  is  the  segment  divided,  and  the 
letters  C  and  D  are  considered  as  being  interchangeable  : 
CDAB,     ACDB,     CABD,     ABCD. 

T   ,  AC.BD,  T,,        AC       J  AD  , 

2.  Let  ^^-g^  be  -.     Then    —   and  --^  have  opposite 

signs,  and  one  point  divides  AB  internally  and  the  other 
externally. 
The  order  of  the  points  is  then  one  of  the  set  CADB,  ACBD. 

3.  When  either  of  the  two  anharmonic  ratios  is  ±1,  these 
ratios  are  equal. 

^  ^'  ^?=  +  ■•    '^''^"  ^=nG'  ^"-^  ^  ^"^  °  ^^« 

both  internal  or  both  external. 


254  synthetic  geometry. 

.,  ac-bc   ad-bd      ab    ab 

^^^°  bc-=Td-'"'bc=bd' 

and  C  and  D  coincide. 

Hence,  when  C  and  D  are  distinct  points,  the  anharmonic 
ratio  of  the  parts  into  which  C  and  D  divide  AB  cannot  be 
positive  unity. 

^    .    .  AC.BD         ^      T,,    „  AC        AD 

5-  L^^AD7BC=-^-     ^^^^BC=-BD- 

And  since  C  and  D  are  now  one  external  and  one  internal 
(2),  they  divide  the  segment  AB  in  the  same  ratio  internally 
and  externally,  disregarding  sign.  Such  division  of  a  line 
segment  is  called  harmonic.  (208°,  Cor.  i) 

Harmonic  division  and  harmonic  ratio  have  been  long  em- 
ployed, and  from  being  only  a  special  case  of  the  more 
general  ratio,  this  latter  was  named  "anharmonic"  by 
Chasles,  "  who  was  the  first  to  perceive  its  utility  and  to 
apply  it  extensively  in  Geometry." 

300°.  Def. — When  we  consider  AB  and  CD  as  being  two 
segments  of  the  same  line  we  say  that  CD  divides  AB,  and 
that  AB  divides  CD. 

Now  the  anharmonic  ratios  in  which  CD  divides  AB  are 
AC.BD        ,  AD  .  BC 
AD.BC  ^"     AC.BD' 
And  the  anharmonic  ratios  in  which  AB  divides  CD  are 
CA^DB        ,  CB.DA 
CB.DA  CA.DB* 

But  the  anharmonic  ratios  of  these  sets  are  equal  each  to 
each  in  both  sign  and  magnitude. 

.'.  thea7iharmonic  ratios  in  which  CD  divides  AB  are  the 
same  as  those  in  which  AB  divides  CD. 

Or,  any  two  segments  of  a  common  tine  divide  each  other 
eqiiianharmonically. 

301°.  Four  points  A,  B,  C,  D  taken  on  a  line  determine  six 
segments  AB,  AC,  AD,  BC,  BD,  and  CD. 


HARMONIC  AND  ANHARMONIC  DIVISION.     255 

These  may  be  arranged  in  three  groups  of  two  each,  so  that 
in  each  group  one  segment  may  be  considered  as  dividing 
the  others,  viz.,     AB,  CD  ;  BC,  AD  ;  CA,  BD. 

Each  group  gives  two  anharmonic  ratios,  reciprocals  of  one 
another ;  and  thus  the  anharmonic  ratios  determined  by  a 
range  of  four  points,  taken  in  all  their  possible  relations,  are 
six  in  number,  of  which  three  are  reciprocals  of  the  other 
three. 

These  six  ratios  are  not  independent,  for,  besides  the 
reciprocal  relations  mentioned,  they  are  connected  by  three 
relations  which  enable  us  to  find  all  of  them  when  any  one  is 
given. 

P,^„_AC.BD,     p   BA.CD,      ^    CB.AD,     ^ 

^^""'^  ADTBC  ^^  ^'  BD7CA  ^^  ^'  CDTAB  ^^  ^^ 

Then  P,  O,  R  are  the  anharmonic  ratios  of  the  groups 
ABCD,  BCAD,  and  CABD,  each  taken  in  the  same  order. 

But  in  any  range  of  four  (233°)  we  have 

AB.CD  +  BC.AD  +  CA.BD=o. 
And  dividing  this  expression  by  each  of  its  terms  in  succes- 
sion, we  obtain  Q  +  ^^  =  R  +  ^  =  P  +  -  =  i. 

From  the  symmetry  of  these  relations  we  infer  that  any 
general  properties  belonging  to  one  couple  of  anharmonic 
ratios,  consisting  of  any  ratio  and  its  reciprocal,  belong 
equally  to  all. 

Hence  the  properties  of  only  one  ratio  need  be  studied. 

The  symbolic  expression  {ABCD}  denotes  any  one  of  the 
anharmonic  ratios,  and  may  be  made  to  give  all  of  them  by 
reading  the  constituent  letters  in  all  possible  orders. 

Except  in  the  case  of  harmonic  ratio,  or  in  other  special 
cases,  we  shall  read  the  symbol  in  the  one  order  of  alternat- 
ing the  letters  in  the  numerator  and  grouping  the  extremes 
and  means  in  the  denominator.     Thus 

{ABCD}  denotes  ^§^- 

It  is  scarcely  necessary  to  say  that  whatever  order  may  be 


256  SYNTHETIC   GEOMETRY. 

adopted  in  reading  the  symbol,  the  same  order  must  be  em- 
ployed for  each  when  comparing  two  symbols. 

302°.  Theorem. — Any  two  constituents  of  the  anharmonic 
symbol  may  be  interchanged  if  the  remaining  two  are  inter- 
changed also,  without  affecting  the  value  of  the  symbol. 

Proof.-        {ABCD}  =  AC.BD  :  AD.BC. 
Interchange  any  two  as  A  and  C,  and  also  interchange  the 
remaining  two  B  and  U.     Then 

{CDAB}=CA.DB:CB.DA 
=  AC.BD  :  AD.BC. 
Similarly  it  is  proved  that 

[ABCD}  =  {BADC}  =  {CDAB}  =  {DCBA}.  q.e.d. 

303°.  If  interchanging  the  first  two  letters,  or  the  last  two, 
without  interchanging  the  remaining  letters,  does  not  alter 
the  value  of  the  ratio,  it  is  harmonic. 

For,  let  {ABCD}  =  {ABDC}. 

^,  AC.BD_AD^BC 

AD.BC    AC.BD' 
or,  multiplying  across  and  taking  square  roots, 

AC.  BD=±  AD.BC. 
But  the  positive  value  must  be  rejected  (299°,  4),  and  the 
negative  value  gives  the  condition  of  harmonic  division. 

304".  Let  ABCD  be  any  range  of  four  and  O  any  point  not 

on  its  axis. 

The  anharmonic  ratio  of  the  pencil 

O .  ABCD  corresponding  to  any  given 

ratio  of  the  range  is  the  same  function 

of  the  sines  of  the  angles  as  the  given 

ratio  is  of  the  corresponding  segments. 

ry,^^    sin  AOC.  sin  BOD  ,    ,     AC.BD 

Thus   -.  - — •    tt^^  corresponds  to   ,  ,^     -^  ; 

smAOD.smBOC  ^  AD.BC 

r,  symbolically,  0{ABCD}  corresponds  to  {ABCD}. 


HARMONIC  AND  ANHARMONIC  DIVISION.     257 

To  prove  that  the  corresponding  anharmonic  ratios  of  the 

range  and  pencil  are  equal. 

AC_AAOC_OA.OCsinAOC_OA   sinAOC 
BC     ABOC     OB.OCsinBOC     OB '  sin  BOC 

o.    .,    ,  BD     OB    sin  BOD 

Similarly,  aD  =  OA' ilKAOD' 

AC  .  BD  _  sin  AOC .  sin  BOD 
AD.BC     sinAOD.sinBOC* 
Hence,  symbolically 

{ABCD}=0{ABCD}  ; 
and,  with  necessary  formal  variations,  the  anharmonic  ratio 
of  a  range  may  be  changed  for  that  of  the  corresponding 
pencil,-  and  vice  versa,  whenever  required  to  be  done. 

Cor.  I.  Two  angles  with  a  common  vertex  divide  each 
other  equianharmonically.  (300°) 

Cor.  2.  If  the  anharmonic  ratio  of  a  pencil  is  +1,  two  rays 
coincide,  and  if  - 1,  the  pencil  is  harmonic.  (299°,  4,  5) 

Cor.  3.  A  given  range  determines  an  equianharmonic  pen- 
cil at  every  vertex,  and  a  given  pencil  determines  an  equian- 
harmonic range  on  every  transversal. 

Cor.  4.  Since  the  sine  of  an  angle  is  the  same  as  the  sine 
of  its  supplement  (214°,  i),  any  ray  may  be  rotated  through  a 
straight  angle  or  reversed  in  directionwithout  affecting  the  ratio. 

Corollaries  2,  3,  and  4  are  of  special  importance. 

305.  Theorem, — If  three  pairs  of  corresponding  rays  of  two 
equianharmonic  pencils  intersect  collinearly,  the  fourth  pair 
intersect  upon  the  line  of  collinearity. 

Proof.— \.Q.\. 

0{ABCD}=0'{ABCD'}, 
and  let  the  pairs  of  corresponding  rays 
OA  and  O'A,  OB  and  O'B,  OC  and  O'C 
intersect  in  the  three  collinear  points  A, 
B,  and  C.  Let  the  fourth  corresponding 
rays  meet  the  axis  of  ABC  in  D  and  D' 
respectively.     Then  {ABCD}  =  ^ABCD'},   (304°) 

R 


258  SYNTHETIC  GEOMETRY. 

AD.BC     AD'.BC'^"     AD     AD" 
which  is  possible  only  when  D  and  D'  coincide. 

.*.  the  fourth  intersection  is  upon  the  axis  of  A,  B,  and  C, 
and  the  four  intersections  are  collinear.  q.e.d. 

Cor.  If  two  of  the  corresponding  rays  as  OC  and  0"C 
become  one  line,  these  rays  may  be  considered  as  intersecting 
at  all  points  on  this  line,  and  however  A  and  B  are  situated 
three  corresponding  pairs  of  rays  necessarily  intersect  col- 
linearly. 

.*.  when  two  equianharmonic  pencils  have  a  pair  of  cor- 
responding rays  in  common,  the  remaining  rays  intersect 
collinearly. 


306°.  Theoretn. — If  two  equianharmonic  ranges  have  three 
pairs  of  corresponding  points  in  perspective,  the  fourth 
points  are  in  the  same  perspective. 

Proof.— 

{ABCD}  =  {A'B'C'D'}, 
and  A  and  A',  B  and  B',  and  C  and  C 
are  in  perspective  at  O.     Now 

0{ABCD}  =  0{A'B'C'D'}, 
and  we  have  two  equianharmonic  pen- 
cils of  which  three  pairs  of  correspond-  ^' 
ing  rays  meet  collinearly  at  A,  B,  and  C.     Therefore  OD' 
and  OD  meet  at  D,  or  D  and  D'  are  in  perspective  at  O. 

Cor.  If  two  of  the  corresponding  points,  as  C  and  C",  be- 
come coincident,  these  two  points  are  in  perspective  at 
every  centre,  and  hence  three  corresponding  pairs  of  points 
are  necessarily  in  perspective. 

.*.  when  two  equianharmonic  ranges  have  a  pair  of  cor- 
responding points  coincident,  the  remaining  pairs  of  cor- 
responding points  are  in  perspective. 


HARMONIC   RATIO.  259 

SECTION    II. 

HARMONIC   RATIO. 

307".  Harmonic  ratio  being  a  special  case  of  anharmonic 
ratio  (299°,  5),  the  properties  and  relations  of  the  latter 
belong  also  to  the  former. 

The  harmonic  properties  of  a  divided  segment  may  ac- 
cordingly be  classified  as  follows  : — 

1.  The  dividing  points  alternate  with  the  end  points  of  the 
divided  segment. 

For  this  reason  harmonic  division  is  symbolized  by  writing 
the  letters  in  order  of  position,  as,  {APBO},  where  A  and  B 
are  the  end  points  of  the  segment  and  P  and  Q  the  dividing 
points  (301°).  A— P— B— O. 

2.  The  dividing  points  P  and  O  divide  the  segment  extern- 
ally and  internally  in  the  same  ratio,  neglecting  sign.  (299°,  5) 

3.  If  one  segment  divides  another  harmonically,  the  second 
also  divides  the  first  harmonically.  (300°) 

4.  A  harmonic  range  determines  a  harmonic  pencil  at  every 
vertex,  and  a  harmonic  pencil  determines  a  harmonic  range 
on  every  transversal.  (304°^  Cor.  3) 

5.  If  one  or  more  rays  of  a  harmonic  pencil  be  reversed  in 
direction  the  pencil  remains  harmonic.  (304°)  Cor.  4) 

6.  Two  harmonic  pencils  which  have  three  pairs  of  corre- 
sponding rays  intersecting  collinearly  have  all  their  corre- 
sponding rays  intersecting  collinearly.  (305°) 

7.  Two  harmonic  ranges  which  have  three  pairs  of  corre- 
sponding points  in  perspective  have  all  their  corresponding 
points  in  perspective.  (306°) 

8.  If  two  harmonic  pencils  have  a  corresponding  ray  from 
€ach  in  common,  all  their  corresponding  rays  intersect  col- 
linearly. (305'*  Cor.) 

9.  If  two  harmonic  ranges  have  a  corresponding  point 
from  each  in  common,  all  their  corresponding  points  are  in 
perspective.  (306°,  Cor.) 


26o  SYNTHETIC   GEOMETRY. 

aoS"".  Let  APBO  be  a  harmonic  range.     Then 
-I  ,      .  .  AP:PB  =  AQ:BQ, 

^  ^    ^        ^    :.      AP:AO  =  AB-AP:AO-AB. 

Taking  AP,  AB,   AG    as   three  magnitudes,  we  have  the 
statement  : — 

The  first  is  to  the  third  as  the  difference  between  the  first 
and  second  is  to  the  difference  between  the  second  and  the 
third.  And  this  is  the  definition  of  three  quantities  in 
Harmonic  Proportion  as  given  in  Arithmetic  and  Algebra. 

Exercises. 

1.  When  three  line  segments  are  in  harmonic  proportion  the 

rectangle  on  the  mean  and  the  sum  of  the  extremes  is 
equal  to  twice  the  rectangle  on  the  extremes. 

2.  The    expanded    symbol    {APBQ}=-i    gives   AP  :  AG 

=  -  BP  :  BQ.     Why  the  negative  sign  ? 

3.  Prove  from  the  nature  of  harmonic  division  that  when  P 

bisects  AB,  Q  is  at  00 . 

4.  Prove  that  if  OP  bisects  z_AOB  internally  GO  bisects  it 

externally  ;  O  {APBQ}  is  equal  to  —  i. 

5.  Trace  the  changes  in  the  value  of  the  ratio  AC  :  BC  as  G 

moves  from  -  00  to  +  00 . 

309*.  In  the  harmonic  range  APBQ,  P  and  Q  are  called 
conjugate  points,  and  so  also  are  A  and  B. 

Similarly  in  the  harmonic  pencil  O.APBQ,  OP  and  OQ 
are  conjugate  rays,  and  so  also  are  OA  and  OB. 

Ex.  I.  Given  three  points  of  a  harmonic  range  to  find  the 
fourth. 

Let  A,  P,  B  be  the  three  given  points. 

By  (259°,  Ex.  7)  find  any  point  O  at  which  the  segments 
AP  and  PB  subtend  equal  angles.  Draw  00  the  external 
bisector  of  the  ^AOB.     Q  is  the  fourth  point. 

For  OP  and  OQ  are  internal  and  external  bisectors  of  the 
,z_AOB.  (208",  Cor.  i) 


HARMONIC    RATIO. 


261 


Ex.  2.  Given  three  rays  to  find  a  fourth  so  as  to  make  the 
pencil  harmonic.  o 

Let  OA,  OP,  OB  be  the  three  rays.  0^ 

'On  OA  take  any  two  equal  distances       ^^ 
OD  and  DE. 

Draw  DF  |i  to  OB,  and  draw  OQ  ||  to 
EF.     OQ  is  the  fourth  ray  required. 

For  since  OD  =  DE,  EF  =  FG.     And  OQ  meets  EF  at  00 


are    harmonic    and    hence    O  .  APBQ    are 


Then   EFGoo 
harmonic. 

Cor.  In  the  symbolic  expression  for  a  harmonic  ratio  a  pair 
of  conjugates  can  be  interchanged  without  destroying  the 
harmonicism. 

{APBQ}  =  {BPAQ}  =  {BQAP}  =  {AQBP}, 
for  {APBQ}  gives  AP  .  BQ  :  AG  .  BP=  -  i, 

and         {BPAQ}  gives  BP  .  AG  :  BQ  .  AP, 
and  being  the  reciprocal  of  the  former  its  value  is  -  i  also. 

And  similarly  for  the  remaining  symbols. 


HARMONIC  PROPERTIES  OF  THE  TETRAGRAM 
OR   COMPLETE    QUADRANGLE. 

310°.  Let  ABGD  be  a  quadrangle,  of  which  AC,  BD,  and 
EF  are  the  three  diagonals.  (247°,  Def  2) 

Also  let  the  line  EO  cut  two  sides 
in  G  aijd  H,  and  the  line  FO  cut 
the  other  two  sides  in  K  and  L. 
Then    . 

I.  AEDisa  Awhereof  AC,  EH, 

and  DB  are  concurrent  lines  from 

the  vertices  to  the  opposite  sides. 

.-.   AB.EC.DH=-AH.DC.EB. 

(251°,  b) 

Also,  AED  is  a  A  and  FCB  is  a 
transversal. 

AB.EC.DF  =  AF.DC 


262  SYNTHETIC   GEOMETRY. 

and  dividing  the  former  equality  by  the  latter, 

DH__AH 

DF         AF' 
and  AHDF  is  a  harmonic  range. 

2.  Again,  {AHDF}  =  E{AHDF}  =  E{LOKF}  =  E{BGCF}. 

(307°,  4) 
LOKF  and  BGCF  are  harmonic  ranges. 

3.  0{AHDF}=0{CEDK}  =  F{CEDK}  =  F{GEHO} 

=  F{BEAL},  (307°,  4,  5) 

but       {CEDK}  =  {DKCE},  etc.  (309°,  Cor.) 

DKCE,  HOGE,  ALBE  are  harmonic  ranges. 

4.  If  AC  be  produced  to  meet  EF  in  I,  AOCI  is  a  harmonic 
range. 

.'.  all  the  lines  upon  which  four  points  of  the  figure  lie  are 
divided  harmonically  by  the  points. 

And  the  points  E,  F,  and  O  at  which  four  hnes  concur  are 
vertices  of  harmonic  pencils. 

Exercises. 

1.  A  line  ||  to  the  base  of  a  A  has  its  points  of  intersection 

with  the  sides  connected  transversely  with  the  end 
points  of  the  base.  The  join  of  the  vertex  with  the 
point  of  intersection  of  these  connectors  is  a  median, 
and  is  divided  harmonically. 

(Let  F  go  to  00  in  the  last  figure.) 

2.  ABC  is  a  A  and  BD  is  an  altitude.     Through  any  point 

O  on  BD,  CO  and  OA  meet  the  sides  in  F  and  E 
respectively.  Show  that  DE  and  DF  make  equal 
angles  with  AC. 

3.  The  centres  of  two  circles  and  their  centres  of  similitude 

form  a  harmonic  range. 

4.  In  the  Fig.  of  310°  the  joins  DI,  IB,  BH,  and  LD  are  all 

divided  harmonically. 


HARMONIC   RATIO.  263 

311°.  Let  APBQ  be  a  harmonic  range  and  let  C  be  the 

middle  point  of  AB.     Then 

AP__BP_PB 

AQ        BQ    BQ- 

T.,,,  •    CB  +  CP_CB-CP 

^^^^  'CB  +  CQ"CQ-CB» 

CB+CP_CQ+CB 

°^  CB-CP     CQ-CB' 

,  CB     CO 

whence  ^=^. 

.*.  CP .  CQ  =  CB2,  Qj.  p  aj^fj  Q  aj.g  inverse  points  to  the  circle 
having  C  as  centre  and  CB  as  radius. 

.'.  I.  The  diameter  of  a  circle  is  divided  harmonically  by 
any  pair  of  inverse  points. 

And  a  circle  having  a  pair  of  co7ijugates  of  a  harmonic 
range  as  end-points  of  a  diameter  has  the  other  pair  of 
conjugates  as  inverse  points. 

Again,  let  EF  be  any  secant  through  P  meeting  the  polar 
of  P  in  V. 

A  circle  on  PV  as  diameter  passes  through  Q  and  P,  and 
therefore  cuts  S  orthogonally.  (258°) 

Hence  also  the  circle  S  cuts  the  circle  on  PV  orthogonally, 
and  E  and  F  are  inverse  points  to  the  circle  on  PV. 

.'.  EPFV  is  a  harmonic  range. 

.*.  2.  A  line  is  cut  harmonically  by  a  pointy  a  circle^  and 
the  polar  of  the  point  with  respect  to  the  circle. 

Ex.  P,  Q  are  inverse  points,  and  from  Q  a  line  is  drawn 
cutting  the  circle  in  A  and  B.  The  join  PB  cuts  the  circle  in 
A'.     Then  AA'  is  ±  to  PQ. 

312°.  Let  P  be  any  point  and  L  its  polar  with  respect  to 
the  circle  Z,  And  let  PCD  and  PBA  be  any  two  secants. 
Then 

I.  PCED  and  PBFA  are  harmonic  ranges  having  P  a  cor- 
responding point  in  each.  Therefore  AD,  FE,  and  BC  arc 
concurrent.     And  BC  and  AD  meet  on  the  polar  of  P.   (309°,  9) 


264 


SYNTHETIC   GEOMETRY. 


2.  Again,  since        {PCED}={PDEC},  (309°,  Cor.) 

.*.  PDEC  and  PBFA  are  harmonic  ranges  having  P  a  cor- 
responding point  in  each.  Therefore  DB,  EF,  and  CA  are 
concurrent,  and  AC  and  DB  meet  on  the  polar  of  P. 


.*.  If  from  any  point  two  secants  be  dj'aivft  to  a  circle,  the 
connectors  of  their  points  of  intersection  with  the  circle  meet 
upon  the  polar  of  the  first  point. 

3.  Since  O  is  on  the  polar  of  P,  P  is  on  the  polar  of  O. 
But  since  Q  is  a  point  from  which  secants  are  drawn  satis- 
fying the  conditions  of  2,  Q  is  on  the  polar  of  O. 

.-.  PQ  is  the  polar  of  O. 

Now  ABCD  is  a  concyclic  quadrangle  whereof  AC  and 
BD  are  internal  diagonals  and  PQ  the  external  diagonal. 

.*.  In  any  concyclic  qicadrangle  t}ie  external  diagonal  is  the 
polar  of  the  point  of  intersection  of  the  internal  diagoiials, 
with  respect  to  the  circumcircle. 

4.  Since  Q  is  on  the  polar  of  P  and  also  on  that  of  O, 
therefore  PO  is  the  polar  of  O,  and  POQ  is  a  triangle  self- 
conjugate  with  respect  to  the  circle. 

5.  Let  tangents  at  the  points  A,  B,  C,  D  form  the  circum- 
scribed quadrangle  USVT. 


HARMONIC   RATIO. 


265 


Then  S  is  the  pole  of  AB,  and  T  of  DC. 
.*.  ST  is  the  polar  of  P,  and  S  and  T  are  points  on  the 
line  QO. 

Similarly  U  and  V  are  points  on  the  line  PC. 

But  XY  is  the  external  diagonal  of  USVT,  and  its  pole  is 

0.  the  point  of  intersection  of  DB  and  AC. 
.*.  X  and  Y  are  points  on  the  line  PQ. 

Hence,  If  tatigents  be  drazun  at  the  vertices  of  a  coney  cite 
quadrangle  so  as  to  form  a  circninscribed  quadrangle^  the  in- 
ternal diagonals  of  the  two  quadrangles  are  concurrent,  and 
their  external  diagonals  are  seginents  of  a  coinmon  line;  and 
the  point  of  concurrence  and  the  line  are  pole  and  polar  with 
respect  to  the  circle. 

Exercises. 

1.  UOVP  and  SOTQ  are  harmonic  ranges. 

2.  If  DB  meets  the  line  PQ  in  R,  lOR  is  a  self-conjugate 

triangle  with  respect  to  the  circle. 

3.  To  find  a  circle  which  shall  cut  the   sides   of  a  given 

triangle  harmonically. 

4.  QXPY  is  a  harmonic  range. 

313°.  Let  S  be  a  circle  and  A'P'B'Q'  a  harmonic  range. 

Taking    any    point    O   on   the  p 

•circle  and  through  it  projecting 
rectilinearly  the  points  A'P'B'Q' 
we  obtain    the    system   APBO, 

which  is  called  a  harmonic  sys-  q,^^ \"""T\~~ \ ^~A^ 

tern  of  points  on  the  circle. 

Now,  taking  O'  any  other  point 
on  the  circle,  O'.  APBQ  is  also  V     \  \  I   /      y^' 

harmonic.     For 

^AOP=_AO'P, 
aPOB  =  ^PO'B,  etc. 
.'.  Def — Four  points  on  a  circle 

formaharmonicsystem  when  their     A  p     b'        q' 

joins  with  any  fifth  point  on  the  circle  form  a  harmonic  pencil. 


266  SYNTHETIC   GEOMETRY. 

Cor.  I.  Since    sinz.AOP  =  i^^,  sin^POB  =  £-^,  etc.,  (228°) 
a  a 

.-.  (304°),  neglecting  sign,  AP .  BQ  =  AQ  .  PB, 

.•.  When  four  poiiits  form  a  harmonic  system  on  a  circle^  the 

rectangles  on  the  opposite  sides  of  the  nor7nal  quadrangle 

which  they  determine  are  equal. 

Cor.  2.  If  O  comes  to  A,  the  ray  OA  becomes  a  tangent 
at  A. 

.*.  When  fottr poi7its  form  a  harmonic  system  on  a  circle^ 
the  tangent  at  any  one  of  them  and  the  chords  from  the  point 
of  contact  to  the  others  form  a  harmo7iic  pencil. 

314°.  Let  the  axis  of  the  harmonic  range  APBQ  be  a  tan- 
gent to  the  circle  S. 

Through  A,  P,  B,  and 
Q  draw  the  tangents  A^,. 
B^,  Vp,  and  Q^. 

These  four  tangents 
form  a  harmonic  system 
of  tangents  to  the  circle  S- 
Let  L  be  any  other  tan- 
gent cutting  the  four  tan- 
gents of  the  system  in  A' 
P',  B',  and  (^. 

Then,  considering  A«,  P^,  B(5,  etc.,  as  fixed  tangents,  and 
A'P'B'Q'  as  any  other  tangent. 

lA.OV  =  lA.O'V\  z.POB  =  ^P'OB',  etc.,  (116°,  Ex.  i> 
.-.  the  pencils  O.APBQ  and  O .  A'P'B'Q'  are  both  har- 
monic, and  A'P'B'Q'  is  a  harmonic  range. 

.*.  Whetifour  tangents  form  a  harjnonic  system  to  a  circle, 
they  intersect  any  other  tangent  in  points  which  foi'i7i  a  har- 
7no7iic  range. 

Cor.  I.  If  the  variable  tangent  coincides  with  one  of  the 
fixed  tangents,  the  point  of  contact  of  the  latter  becomes  one 
of  the  points  of  the  range. 


HARMONIC   RATIO.  26/ 

.'.  When  four  ta7igenis  form  a  harmo7iic  system  to  a  circle, 
each  tangent  is  divided  harmonically  by  its  point  of  contact 
and  its  intersections  with  the  other  tangents. 

Exercises. 

1.  Tangents  are  drawn  at  A,  A'  the  end-points  of  a  diameter, 

and  two  points  P,  B  are  taken  on  the  tangent  through 
A  such  that  AB  =  2AP.  Through  P  and  B  tangents  are 
drawn  cutting  the  tangent  at  A'  in  P'  and  B'.  Then 
2A'B'  =  A'P',  and  AA',  PB',  and  BP'  are  concurrent. 

2.  Four  points  form  a  harmonic  system  on  a  circle.     Then  the 

tangents  at  one  pair  of  conjugates  meet  upon  the  secant 
through  the  other  pair. 

3.  If  four  tangents  form  a  harmonic  system  to  a  circle,  the 

point  of  intersection  of  a  pair  of  conjugate  tangents  lies 
on  the  chord  of  contact  of  the  remaining  pair. 

4.  If  four  points  form  a  harmonic  range,  their  polars  with 

respect  to  any  circle  form  a  harmonic  pencil ;  and 
conversely. 


SECTION   III. 

OF    ANHARMONIC   PROPERTIES. 

315°.  Let  A,  E,  C  and  D,  B,  F  be  two  sets  of  three  col- 
linear  points  having  their  ^ 
axes     meeting     in     some 
point  R. 

Join  the  points  alter- 
nately, as  ABCDEFA. 
Then  AB  and  DE,  BC  and 
EF,  CD  and  FA  meet  in 
P,  Q,  O.    To  show  that  these  points  are  collinear. 


268  SYNTHETIC   GEOxMETRY. 

0{ECQF}  =  C{EOQF}  (referred  to  axis  EF) 

=  C{RDBF}  (referred  to  axis  DR) 
-A{RDBF} 

=  A{EDPF}  (referred  to  axis  DE) 
=  0{EDPF} 

=  0{ECPF}.  (by  reversing  rays,  etc.) 

.  .'.  the  pencils  O.  ECQF  and  O.  ECPF  are  equianharmonic, 
and  having  three  rays  in  common  the  fourth  rays  must  be  in 
common,  /.<?.,  they  can  differ  only  by  a  straight  angle,  and 
therefore  O,  P,  Q  are  collinear. 

(Being  the  first  application  of  anharmonic  ratios  the  work 
is  very  much  expanded.) 

.' .  If  six  lines  taken  in  order  intersect  alternately  in  two 
sets  of  three  collinear  points^  they  intersect  in  a  third  set  of 
three  collijiear  points. 

Cor.  I.  ABC  and  DEF  are  two  triangles,  whereof  each  has 
one  vertex  lying  upon  a  side  of  the  other: 

If  AB  and  DE  are  taken  as  corresponding  sides,  A  and  F 
are  non-corresponding  vertices.  But,  if  AB  and  EF  are 
taken  as  corresponding  sides,  A  and  D  are  non-corresponding 
vertices. 

Hence  the  intersections  of  AB  and  EF,  of  ED  and  CB, 
and  of  AD  and  CF  are  collinear. 

.*.  If  two  triangles  have  each  a  vertex  lying  upon  a  side  of 
the  other,  the  remaining  sides  and  the  joins  of  the  remaining 
non-corresponding  vertices  intersect  collinearly. 

Cor.  2.  Joining  AD,  BE,  CF,  ADBE,  EBFC,  and  ADFC 
are  quadrangles,  and  P,  O,  O  are  respectively  the  points  of 
intersection  of  their  internal  diagonals. 

.*.  if  a  quadrangle  be  divided  into  two  quadrangles,  the 
points  of  intersection  of  the  internal  diagonals  of  the  three 
quadrangles  are  collinear. 


OF   ANHARMONIC   PROPERTIES. 


269 


316",  Let  A,  A',  B,  B',  C,  C  be  six  points  lying  two  by  two 
on  two  sets  of  three  con- 
current lines,  which  meet 
at  P  and  O.  Then  the 
points  lie  upon  a  third  set 
of  three  concurrent  lines 
meeting  at  O. 

We  are  to  prove  that 
AG  and  AA'  are  in  line. 
A{0,rC'B}  =  B'{0:irC'Q} 

=  B'{CPC>} 

=  Q{A'P^B} 

=A{A';rC'B}. 
.'.  the  pencils  A.OjtC'B 
and  A.A';i'C'B   are  equi- 
anharmonic,  and  have  three  corresponding  rays  in  common. 
Therefore  AG  and  A  A'  are  in  line. 

Cor.  ABC  and  A'B'C  are  two  As  which  are  in  perspective 
at  both  P  and  Q,  and  we  have  shown  that  they  are  in  per- 
spective at  G  also. 

As  there  is  an  axis  of  perspective  corresponding  to 
each  centre,  the  joins  of  the  six  points,  accented  letters 
being  taken  together  and  unaccented  together,  taken  in 
every  order  intersect  in  three  sets  of  three  collinear 
points. 


Exercises. 

If  two  As  have  their  sides  intersecting  collinearly,  their 
corresponding  vertices  connect  concurrently. 

The  converse  of  Ex.  i. 

Three  equianharmonic  ranges  ABCD,  A'B'C'D',  and 
PQRS  have  their  axes  concurrent  at  Y,  and  AA',  BB', 
CC,  DD'  concurrent  at  X.  Then  the  two  groups  of 
joins  AP,  BO,  CR,  DS,  and  A'P,  B'Q,  C'R,  D'S  are 


2/0 


SYNTHETIC   GEOMETRY. 


concurrent  at  two  points  O  and  O'  which  are  coUinear 
with  X. 

4.  From  Ex.  3  show  that  if  a  variable  A  has  its  sides  passing 

through  three  fixed  points,  and  two  of  its  vertices  lying 
upon  fixed  lines,  its  third  vertex  lies  upon  a  fixed  line 
concurrent  with  the  other  two. 

5.  If  a  variable  A  has  its  vertices  lying  on  three  fixed  lines 

and  two  of  its  sides  passing  through  fixed  points,  its 
third  side  passes  through  a  fixed  point  coUinear  with 
the  other  two. 


The  range  ABCD  is  transferred  to  the  circle  S  by 
rectilinear  projection  through 
any  point  O  on  the  circle. 
Then  A'B'C'D'  is  a  system  of 
points  on  the  circle  which  is 
equianharmonic  with  the  range 
ABCD. 

I.  If  O'  be  any  other  point 
on  the  circle,  the 

^A'OB'  =  ^A'0'B', 
ABCD  ^B'OC'  =  ^B'0'C',  etc., 

and  the  two  pencils  O .  A'B'C'D'  and  O'.  A'B'C'D'  are  equi- 
anharmonic. 

.'.  four  points  on  a  circle  subtend  equianharmonic  pencils 
at  all  fifth  points  on  the  circle. 


2.  Since 


sinA'OB'  = 


A'B' 


sin  COD': 


etc., 


CD' 

AB .  CD  +  BC .  AD  =  AC .  BD, 
A'B'.  CD'  +  B'C.  A'D'  =  A'C'.  B'D' ; 
which  is  an  extension  of  Ptolemy's  theorem  to  a  concyclic 
quadrangle.  (205°) 


and 


(233°) 


OF  ANHARMONIC   PROPERTIES.  2J\ 

3.  If  the  range  ABCD  is  inverted  with  O  as  the  centre  of 
inversion,  the  axis  of  the  range  inverts  into  a  circle  S  through 
O,  and  A,  B,  C,  D  invert  into  A',  B',  C,  and  D'  respectively. 

Hence,  in  general,  anharmonic  relations  are  unchanged  by 
inversion,  a  range  becoming  an  equianharmonic  system  on  a 
circle,  and  under  certain  conditions  vice  versa. 

4.  In  the  inversion  of  3,  A  and  A',  B  and  B',  etc.,  are  pairs 
of  inverse  points. 

OA.OA'  =  OB.OB' 

=  OC.OC'=OD.OD', 
and  the  As  OAB  and  OB'A',  OAC  and  OCA',  etc.,  are 
similar  in  pairs. 

And  if  P  be  the  ±  from  O  to  AD,  and  P^,  Pg,  P3,  and  P4  be 
the  J_s  from  O  to  A'B',  B'C,  CD',  and  D'A',  we  have 

AB    A'B'     BC     B'C 


P 

Pi' 

P 

P2' 

CD 
P 

CD' 
P3' 

DA 
P  ■ 

D'A' 
P4' 

AB-}-BC  +  CD4 

-DA_ 

=T^ 

B'C  ,  CD' 

"pT^-pr 

P 

D'A' 
„         P4* 
But  (232°)  AB  +  BC  +  CD  +  DA=o, 

A'B' ^  B'C  ,  CD' ^  D'A' 

And  since   the   same  principle  applies  to  a  range  of  any 
number  of  points, 

.*.  in  any  concyclic  polygon,  if  each  side  be  divided  by  the 
perpendicular  upon  it  from  any  fixed  point  on  the  circle,  the 
sum  of  the  quotients  is  zero. 

.  In  the  preceding  theorem,  as  there  is  no  criterion  by  which 
we  can  distinguish  any  side  as  being  negative,  some  of  the 
perpendiculars  must  be  negative. 

Of  the  perpendiculars  one  falls  externally  upon  its  side  of 


2/2 


SYNTHETIC   GEOMETRY. 


the  polygon  and  all  the  others  fall  internally.     Therefore  the 
theorem  may  be  stated  : — 

If  _Ls  be  drawn  from  any  point  on  a  circle  to  the  sides  of  an 
inscribed  polygon,  the  ratio  of  the  side,  upon  which  the  X 
falls  externally,  to  its  ±  is  equal  to  the  sum  of  the  ratios  of 
the  remaining  sides  to  their  _Ls. 


318°.   Theorem. —  If  two  circles  be  inverted  the  ratio  of  the 
square  on  their  common  tangent  to  the  rectangle  on  their 

diameters  is  unchanged. 
Let  S,  S'  be  the 
circles  and  AD  be  the 
common  centre  line, 
and  let  the  circles  s 
and  s'  and  the  circle 
Z  be  their  inverses 
respectively. 

Then  Z  cuts  s  and  / 
orthogonally,  and 
O{ABCDH0{A'B'C'D'}. 
But  if  abed  be  the 
common  centre  line  of 
s  and  s\  aPJ,  bB\  eC, 
and  dl>'  are  concurrent 
at  O.  (265',  Ex.  I) 

0{rt^rrt'}  =  0{A'B'C'D'} 
=  0{ABCD}, 
{«^^^}  =  {ABCD}, 
A.C.BT>_ac.bd 
AB.CD     ab.ed' 

But  AC .  BD  =  the  square  on  the  common  direct  tangent  to  S 
and  S',  and  ac .  (^^=  the  square  of  the  corresponding  tangent 
to  s  and  s'.  (i79°j  Ex.  2) 

And  AB  .  CD  and  ab.ed  diXQ  the  products  of  the  diameters 
respectively. 


or 


OF   ANHARMONIC   PROPERTIES.  2/3 

And  the  theorem  is  proved. 

Cor.  I.  Writing  the  symbolic  expressions  {ABCD}  and 
{adcd}  in  another  form,  we  have 

AD.BC_ad.dc 

AB.CD     ad.cd' 
And  AD .  BC  and  ad.  be  are  equal  to  the  squares  on  the 
transverse  common  tangents  respectively. 

Cor.  2.  If  four  circles  Sj,  S2,  S3,  S4  touch  a  line  at  the 
points  A,  B,  C,  D,  and  the  system  be  inverted,  we  have  four 
circles  s^,  s^^  s^,  ^"4,  which  touch  a  circle  Z  through  the  centre 
of  inversion. 

Now  let  </i,  </2'  ^3»  ^4  be  the  diameters  of  Sj,  S2,  etc.,  and  let 
^i>  ^2>  ^35  ^4  be  the  diameters  of  j^,  s^^  etc.,  and  let  t^^  be  the 
common  tangent  to  s-^  and  s^,  /g^  be  that  to  s^  and  J4,  etc. 

Then  AB,  etc.,  are  common  tangents  to  S^  and  82,  etc.^ 
and  AB.CD  +  BC.AD  +  CA.BD=o.  (233°) 

And 


AB2 
d,d^ 

CD2 

5354' 

AB.CD  ^ 

sld^d^d^^ 

_      ^12^34 

and  similar  equalities  for  the  remaining  terms, 

/l2^34  +  ^23^1 4  +  ^31^4  "  ^• 

This  theorem,  which  is  due  to  Dr.  Casey,  is  an  extension 
of  Ptolemy's  theorem.  For,  if  the  circles  become  point- 
circles,  the  points  form  the  vertices  of  a  concyclic  quadrangle 
and  the  tangents  form  its  sides  and  diagonals. 

If  we  take  the  incircle  and  the  three  excircles  of  a  triangle 
as  the  four  circles,  and  the  sides  of  the  triangle  as  tangents, 
we  obtain  by  the  help  of  Ex.  i,  135°,  (^2  _  ^2  +  ^2  _  ^2  +  ^2  _  ^2 
as  the  equivalent  for  /12/34  +  etc. ;  and  as  this  expression  is 
identically  zero,  the  four  circles  given  can  all  be  touched  by 
a  fifth  circle. 

S 


274 


SYNTHETIC   GEOMETRY. 


319°.  Let  A,  B,  C,  D,  E,  F  be  six  points  on  a  circle  so  con- 
nected as  to  form  a  iiexagram, 
i.e.^  such  that  each  point  is  con- 
nected with  two  others. 

Let  the  opposite  sides  AB, 
DE  meet  in  P  ;  BC,  EF  in  Q ; 
and  CD,  FA  in  R. 

To  prove  that  P,  Q,  and  R  are 
coUinear. 

0{BDER}  =  Q{CDER} 


=  F{CDEA} 
=  B{CDEP} 
=  Q{BDEP}, 
.'.  QR  and  QP  are  in  line. 
.•.    if   a    hexagram    have    its 
vertices  concyclic,  the  points  of 
intersection  of  its  opposite  sides 
in  pairs  are  collinear. 

Dcf. — The  line  of  collinearity  is  called  the  pascal  of  the 
hexagram,  after  the  famous  Pascal  who  discovered  the 
theorem,  and  the  theorem  itself  is  known  as  Pascal's 
theorem. 

Cor.  I.  The  six  points  may  be  connected  in  5x4x3x2  or 
120  different  ways.  For,  starting  at  A,  we  have  five  choices 
for  our  first  connection.  It  having  been  fixed  upon,  we  have 
four  for  the  next,  and  so  on  to  the  last.  But  one-half  of  the 
hexagrams  so  described  will  be  the  other  half  described  by 
going  around  the  figure  in  an  opposite  direction.  Hence,  six 
points  on  a  circle  can  be  connected  so  as  to  form  60  different 
hexagrams.  Each  of  these  has  its  own  pascal,  and  there  are 
thus  60  pascal  lines  in  all. 

When  the  connections  are  made  in  consecutive  order  about 
the  circle  the  pascal  of  the  hexagram  so  formed  falls  without 
the  circle  ;  but  if  any  other  order  of  connection  is  taken,  the 
pascal  may  cut  the  circle. 


OF   ANIIARMONIC   PROPJiRTIKS.  2/5 

Cor  2.  In  the  hexagram  in  the  figure,  the  pascal  is  the  line 
through  P,  O,  R  cutting  the  circle  in  H  and  K.     Now 
C{KFBD}  =  C(KFQR}  ^ — 

=  F{KCOR}  a//\       \c 

=  F{KCEA},  /W     \yf\ 

{KFBD}={KCEA},  (    \j\)W     j 

and  K  is  a  common  point  to  two  equi-    hV  ^/V/^SAP   J^ 
anharmonic  systems  on  the  circle.     So  xTX  /     ^xf 

also  is  H.  B 

These  points  are  important  in  the  theory  of  homographic 
systems. 

Cor.  3.  Let  i,  2,  3,  4,  5,  6  denote  six  points  taken  consecu- 
tively upon  a  circle.  Then  any  particular  hexagram  is  denoted 
by  writing  the  order  in  which  the  points  are  connected,  as  for 
example,  2461352. 

In  the  hexagram  246135  the  pairs  of  opposite  sides  are  24 
and  13,  46  and  35,  61  and  52,  and  the  pascal  passes  through 
their  intersections. 

Now  taking  the  four  hexagrams 

246135,  245136,  246315,  2453I6, 
the  pascal  of  each  passes  through  the  intersection  of  the 
connector  of  2  and  4  with  the  connector  of  i  and  3.     Hence 
the  pascals  of  these  four  hexagrams  have  a  common  point. 

It  is  readily  seen  that  inverting  the  order  of  2  and  4  gives 
hexagrams  which  are  only  those  already  written  taken  in  an 
inverted  order. 

.•.  the  pascals  exist  in  concurrent  groups  of  four,  meeting 
at  fifteen  points  which  are  intersections  of  connectors. 

Cor.  4.  In  the  hexagram  1 352461  consider  the  two  triangles 
formed  by  the  sides  13,  52,  46  and  35,  24,  61.  The  sides  13 
and  24,  35  and  46,  52  and  61  intersect  on  the  pascal  of 
1352461,  and  therefore  intersect  collinearly. 

Hence  the  vertices  of  these  triangles  connect  concurrently, 
i.e.,  the  line  through  the  intersection  of  35  and  61  and  the 
intersection  of  52  and  46,  the  line  through  the  intersection  of 


2/6 


SYNTHETIC   GEOMETRY. 


35  and  24  and  the  intersection  of  I3  and  46,  and  the  line 
through  the  intersections  of  24  and  61  and  the  intersection 
of  13  and  25  are  concurrent. 

But  the  first  of  these  Hnes  is  the  pascal  of  the  hexagram 
1643521,  the  second  is  the  pascal  of  the  hexagram  3564213, 
and  the  third  is  the  pascal  of  the  hexagram  4256134. 

.*.  the  pascals  exist  in  concurrent  groups  of  three,  meeting 
at  20  points  distinct  from  the  1 5  points  already  mentioned. 

Cor.  5.  If  two  vertices  of  the  hexagram  coincide,  the  figure  be- 
comes a  pentagram,  and  the  missing  side  becomes  a  tangent. 

.*,  if  a  pentagram  be  inscribed  in  a  circle  and  a  tangent  at 
any  vertex  meet  the  opposite  side,  the  point  of  intersection 
and  the  points  where  the  sides  about  that  vertex  meet  the 
remaining  sides  are  collinear. 

Ex.  I.  The  tangents  at  opposite  vertices  of  a  concyclic  quad- 
rangle intersect  upon  the  external  diagonal  of  the  quadrangle. 

Ex.  2.  ABCD  is  a  concyclic  quadrangle.  AB  and  CD 
meet  at  E,  the  tangent  at  A  meets  BC  at  G,  and  the  tangent 
at  B  meets  AD  at  F.     Then  E,  F,  G  are  collinear. 


320°.  Let  six  tangents  denoted  by  the  numbers  i,  2,  3,  4,  5, 

and  6  touch  a 
circle  in  A,  B,  C, 
D,  E,  and  F. 
And  let  the 
points  of  inter- 
section of  the 
tangents  be  de- 
noted by  12,  23, 
34,  etc. 

Then  the  tan- 
gents form  a 
hexagram  about 
the  circle. 

Now,  12  is 
the  pole  of  AB,  and  45  is  the  pole  of  ED.     Therefore  the 


OF  ANHARMONIC  PROPERTIES.  277 

line  12.45  is  the  polar  of  the  point  of  intersection  of  AB 
and  ED. 

Similarly  the  line  23 .  56  is  the  polar  of  the  intersection  of 
BC  and  EF,  and  the  line  34.61  is  the  polar  of  the  intersec- 
tion of  CD  and  FA. 

But  since  ABCDEF  is  a  hexagram  in  the  circle,  these 
three  intersections  are  collinear.  (319°) 

.*.  the  lines  12 .  45,  23 .  56,  and  34.  61  are  concurrent  at  O. 

And  hence  the  hexagram  formed  by  any  six  tangents  to  a 
circle  has  its  opposite  vertices  connecting  concurrently. 

Def. — The  point  of  concurrence  is  the  Briaiichon  point,  and 
the  theorem  is  known  as  Brianchoiis  theorei7i. 

Cor.  I.  As  the  six  tangents  can  be  taken  in  any  order  to 
form  the  hexagram,  there  are  60  different  hexagrams  each 
having  its  own  Brianchon  point. 

Now  take,  as  example,  the  hexagram  formed  by  the  lines 
123456  taken  in  order. 

The  connectors  are  12  .  45,  23  .  56,  34.  61,  and  these  give 
the  point  O. 

But  the  hexagrams  126453,  123546,  and  T26543  all  have 
one  connector  in  common  with  123456,  namely,  that  which 
passes  through  12  and  45.  Hence  the  Brianchon  points  of 
these  four  hexagrams  lie  upon  one  connector. 

.*.  the  60  Brianchon  points  lie  in  collinear  groups  of  four 
upon  15  connectors  of  the  points  of  intersection  of  the 
tangents. 

Cor.  2.  Consider  the  triangles  12.56.34  and  45.23.61. 
These  have  their  vertices  connecting  concurrently,  and  there- 
fore they  have  their  sides  intersecting  collinearly. 

But  the  point  of  intersection  of  the  sides  61 .  23  and  56.  34 
is  the  Brianchon  point  of  the  hexagram  formed  by  the  six 
lines  234165  taken  in  order ;  and  similar  relations  apply  to 
the  other  points  of  intersection. 

Hence  the  60  Brianchon  points  lie  in  collinear  groups  of 
three  upon  axes  which  are  not  diagonals  of  the  figure. 


2/8  SYNTHETIC   GEOMETRY. 

Cor.  3.  Let  two  of  the  tangents  become  coincident. 

Their  point  of  intersection  is  then  their  common  point  of 
contact,  and  the  hexagram  becomes  a  pentagram. 

.•.  in  any  pentagram  circumscribed  to  a  circle  the  join  of 
a  point  of  contact  with  the  opposite  vertex  is  concurrent  with 
the  joins  of  the  remaining  vertices  in  pairs. 

Ex.  I.  In  any  quadrangle  circumscribed  to  a  circle,  the 
diagonals  and  the  chords  of  contact  are  concurrent. 

Ex.  2.  In  any  quadrangle  circumscribed  to  a  circle,  the 
lines  joining  any  two  vertices  to  the  two  points  of  contact 
adjacent  to  a  third  vertex  intersect  on  the  join  of  the  third 
and  the  remaining  vertex. 


SECTION   IV. 

OF  POLAR  RECIPROCALS  AND  RECIPROCATION. 

321°.  The  relation  of  pole  and  polar  has  already  been 
explained  and  somewhat  elucidated  in  Part  IV.,  Section  V. 

It  was  there  explained  that  when  a  figure  consists  of  any 
number  of  points,  and  their  connecting  lines,  another  figure 
of  the  same  species  may  be  obtained  by  taking  the  poles  of 
the  connectors  of  the  first  figure  as  points,  and  the  polars  of 
the  points  in  the  first  figure  as  connecting  lines  to  form  the 
second. 

And  as  the  first  figure  may  be  reobtained  from  the  second 
in  the  same  way  as  the  second  is  obtained  from  the  first,  the 
figures  are  said  to  be  polar  reciprocals  of  one  another,  as 
being  connected  by  a  kind  of  reciprocal  relation.  The  word 
reciprocal  in  this  connection  has  not  the  same  meaning  as  in 
184°,  Def. 

The  process  by  which  we  pass  from  a  figure  to  its  polar 
reciprocal  is  called  polar  reciprocation  or  simply  reciprocation. 


OF  POLAR  RECIPROCALS  AND  RECIPROCATION.   279 

322°.  Reciprocation  is  effected  with  respect  to  a  circle 
either  expressed  or  imphed.  The  radius  and  centre  of  this 
reciprocating  circle  are  quite  arbitrary,  and  usually  no 
account  need  be  taken  of  the  radius.  Certain  problems  in 
reciprocation,  however,  have  reference  to  the  centre  of  re- 
ciprocation, although  the  position  of  that  centre  may  gener- 
ally be  assumed  at  pleasure. 

From  the  nature  of  reciprocation  we  obtain  at  once  the 
following  statements  : — 

1.  A  point  reciprocates  into  a  line  and  a  line  into  a  point. 
And  hence  a  figure  consisting  of  points  and  lines  reciprocates 
into  one  consisting  of  lines  and  points. 

2.  Every  rectilinear  figure  consisting  of  more  than  a  single 
line  reciprocates  into  a  rectilinear  figure. 

3.  The  centre  of  reciprocation  reciprocates  into  the  line  at 
00 ,  and  a  centre-line  of  the  circle  of  reciprocation  reciprocates 
into  a  point  at  00  in  a  direction  orthogonal  to  that  of  the 
centre-line. 

4.  A  range  of  points  reciprocates  into  a  pencil  of  lines,  and 
the  axis  of  the  range  into  the  vertex  of  the  pencil.  And 
similarly,  a  pencil  of  lines  reciprocates  into  a  range  of  points, 
and  the  vertex  of  the  pencil  into  the  axis  of  the  range. 


323°.  Let  O.LMNK  be  a  pencil  of  four,  and  C  be  the 
centre     of    reciprocation.        ./ 
Draw  the   perpendiculars 
C/  on  L,  C»z'  on  M,  C«' 
on  N,  and  Qk'  on  K. 

The  poles  of  L,  M,  N, 
K  lie  respectively  on  these 
perpendiculars,  forming  a 
range  of  points  as  /,  m,  11^ 
k.     Then 

I.  Evidently  the 

^LOM  =  ^/C;;/,  ^MON  =  i.wC;/,  etc. 
.*.  the  angle  between  two  lines  is  equal  to  that  subtended  at 


280  SYNTHETIC   GEOMETRY. 

the  centre  of  reciprocation  by  the  poles  of  the  lines  ;  and  the 
angle  subtended  at  the  centre  of  reciprocation  by  two  points 
is  equal  to  the  angle  between  the  polars  of  the  points. 

2.  Any  pencil  of  four  is  equianharmonic  with  its  polar 
reciprocal  range.  And  hence  anharmonic  or  harmonic  re- 
lations are  not  altered  by  reciprocation. 

Def. — Points  are  said  to  be  perpendicular  to  one  another 
when  their  joins  with  the  centre  of  reciprocation  are  at  right 
angles.  In  such  a  case  the  polars  of  the  points  are  perpen- 
dicular to  one  another. 


324°.  In  many  cases,  and  especially  in  rectilinear  figures, 
the  passing  from  a  theorem  to  its  polar  reciprocal  is  quite  a 
mechanical  process,  involving  nothing  more  than  an  intel- 
ligent and  consistent  change  in  certain  words  in  the  statement 
of  the  theorem. 

In  all  such  cases  the  truth  of  either  theorem  follows  from 
that  of  its  polar  reciprocal  as  a  matter  of  necessity. 

Take  as  example  the  theorem  of  88°,  "  The  three  altitudes 
of  a  A  are  concurrent," 

To  get  its  polar  reciprocal  put  it  in  the  following  form,  where 

the  theorem  and  its  polar  reciprocal  are  given  in  alternate 

lines  : — 

rp,  ^ -T^^^^f  lines  through  the  vertices)    r      a  ^•     ^ 

The  three -^      .  ^  ,         .,  ,^of  a  A  Perpendicular 

(pomts    on    the    sides      ) 


to  the  opposite!^'  ^^      [are] 
( vertices )        ( 


concurrent 
collinear. 


To  get  a  point  J_  to  a  vertex  we  connect  the  vertex  to  the 
centre  of  reciprocation,  and  through  this  centre  draw  a  line  X 
to  the  connector.  The  point  required  lies  somewhere  on  this 
line.  (323°,  Def.) 

And  as  the  centre  may  be  any  point,  we  may  state  the 
polar  reciprocal  thus  : — 

"  The  lines  through  any  point  perpendicular  to  the  joins  of 


OF  POLAR  RECIPROCALS  AND  RECIPROCATION.  28 1 

that  point  with  the  vertices  of  a  triangle  intersect  the  opposite 
sides  of  the  triangle  collinearly."  (252°,  Ex.  8) 

325°.  Consider  any  two  As-  These  reciprocate  into  two 
As  ;  vertices  giving  sides,  and  sides,  vertices. 

If  the  original  As  are  in  perspective  their  vertices  connect 
concurrently.  But  in  reciprocation  the  vertices  become  sides 
and  the  point  of  concurrence  becomes  a  line  of  collinearity. 
Hence  the  polar  reciprocals  of  these  As  have  their  sides 
intersecting  collinearly  and  are  in  perspective. 

.*.  As  in  perspective  reciprocate  into  As  in  perspective. 

But  any  three  concurrent  lines  through  the  vertices  of  a  A 
intersect  the  opposite  sides  in  points  which  form  the  vertices 
of  a  new  A  in  perspective  with  the  former. 

Hence  all  cases  of  three  concurrent  lines  passing  through 
the  vertices  of  a  A  reciprocate  into  As  in  perspective  with 
the  original.  Such  are  the  cases  of  the  concurrence  of  the 
three  medians,  the  concurrence  of  the  three  altitudes,  of  the 
three  bisectors  of  the  angles,  etc. 

326°.  The  complete  harmonic  properties  of  the  tetragram 
may  be  expressed  in  the  two  following  theorems,  which  are 
given  in  alternate  lines,  and  are  polar  reciprocals  to  one 
another  : — 

Fourj^'"^'    determine  by  their  j^^^^^^^^^^^^H six  JP^"^^^^ 
(points)  (connectors     )       (lines,    j 

and  the|~""^«?=    iof  these  by  theirl'""='-^^"'°"=  Ideter- 
( mtersections )  ( connectors    ) 

mine  three  new|P°™'^-l     Thel™""^""^    lof  any  of  the 

(hnes.    )  (mtersections) 

three  newjP°'"^^lwith  the  original  six|P''^''^^|form  a  har- 
(hnes    )  (lines    i 

monic  jP^"^^^- 
( range. 

Other  polar  reciprocal  theorems,  which  have  been  already 


282  SYNTHETIC    GEOMETRY. 

given,  are  Pascal's  and  Brianchon's  theorems  with  all  their 
corollaries,  the  theorems  of  Arts.  313"  and  314°,  of  Arts. 
315°  and  316°,  etc. 

The  circle,  when  reciprocated  with  respect  to  any  centre  of 
reciprocation  not  coincident  with  its  own  centre,  gives  rise  to 
a  curve  of  the  same  species  as  the  circle,  ?>.,  a  conic  section, 
and  many  properties  belonging  to  the  circle,  and  particularly 
those  which  are  unaltered  by  reciprocation,  become  properties 
of  the  general  curve. 

These  generalized  properties  cannot  be  readily  understood 
without  some  preliminary  knowledge  of  the  conic  sections. 


SECTION   V. 

HOMOGRAPHY  AND    INVOLUTION. 

327°.  Let  A,  B  and  A',  B'  be  fixed  points  on  two  lines,  and 

• ' ' ' —    let  P  and  P'  be  variable  points, 

BODE  .  . 

___^ one  on  each  line  which  so  move 

B'      C        D'        E'  as  to  preserve  the  relation 


AP_,     AT' 
BP      '  '  B'P" 
where  >6  is  any  constant ;  and  let  C,  C' ;  D,  D' ;  E,  E',  etc., 
be  simultaneous  positions  of  P  and  P'. 

Then  the  points  A,  B,  C,  D,  E,  etc.,  and  A',  B',  C',  D',  E', 
etc.,  divide  homographically  the  lines  upon  which  they  lie. 
AC  _ ;     A'C       1  AD  _  ;     A'D' 
BC-^^-FC'^^'^BD-^'-B'D" 
AC.BD_A^C.B'D^ 
AD  .  BC     A'D' .  B'C" 
or  {ABCD}  =  {A'B'C'D'}. 

Similarly,      {ABCE}  =  {A'B'C'E'},  {BCDE}  =  {B'C'D'E'},  etc. 


HOMOGRAPHY   AND   INVOLUTION.  285 

Evidently  for  each  position  of  P,  P'  can  have  only  one 
position,  and  conversely,  and  hence  the  points  of  division  on 
the  two  axes  correspond  in  unique  pairs. 

.'.  two  lines  are  divided  homographically  by  two  sets  of 
points  when  to  each  point  on  one  corresponds  one  and  only 
one  point  on  the  other,  and  when  any  four  points  on  one  line 
and  their  four  correspondents  on  the  other  form  equianhar- 
monic  ranges. 

Cor.  I.  If  the  systems  of  points  be  joined  to  any  vertices  O 
and  C,  the  pencils  O.ABCD...  and  0'.  A'B'C'D'...  are 
evidently  homographic,  and  cut  all  transversals  in  homo- 
graphic  ranges. 

Cor.  2.  The  results  of  Arts.  304°,  Cors.  3  and  4,  and  of 
Arts.  305°  and  306°  and  their  corollaries  are  readily  extended 
to  homographic  ranges  and  pencils. 

The  following  examples  of  homographic  division  are  given. 

Ex.  I.  A  line  rotating  about  a  fixed  point  in  it  cuts  any  two 
lines  homographically. 

Ex.  2.  A  variable  point  confined  to  a  given  line  determines 
two  homographic  pencils  at  any  two  fixed  points. 

Ex.  3.  A  system  of  ^./.-circles  determines  two  homographic 
ranges  upon  any  Hne  cutting  the  system. 

Consider  any  two  of  the  circles,  let  P,  Q  be  the  common 
points,  and  let  the  line  L  cut  one  of  the  circles  in  A  and  A' 
and  the  other  in  B  and  B'.  Then  the  ^PBB'  =  Z.PQB',  and 
_PAB'  =  ^PQA'.     .-.  £.APB  =  ^A'OB'. 

Hence  the  segment  BA  subtends  the  same  angle  at  P  as 
the  segment  at  B'A'  does  at  O.  And  similarly  for  all  the 
segments  made  in  the  other  circles. 

Ex.  4.  A  system  of  /.^.-circles  determines  two  homographic 
ranges  upon  every  line  cutting  the  system. 


284  SYNTHETIC   GEOMETRY. 


DOUBLE   POINTS  OF  HOMOGRAPHIC   SYSTEMS. 

328°.  Let  ABCD...  and  A'B'C'D'...  be  two  homographic 

^ ^    ^         ^    B  c  p  ranges  on  a  common  axis. 

°  A'      B'        c  D'  If  any  two  correspondents 

from  the  two  ranges  become  coincident  the  point  of  co- 
incidence is  a  double  point  of  the  system. 

If  A  and  A'  were  thus  coincident  we  would  have  the  rela- 
tions {ABCE}  =  {AB'C'E'},  etc. 

Thus  a  double  point  is  a  common  constituent  of  two  equi- 
anharmonic  ranges,  of  which  the  remaining  constituents  are 
correspondents  from  two  homographic  systems  upon  a 
common  axis. 

ABC  and  A'B'C  being  fixed,  let  D  and  D'  be  two  variable 
correspondents  of  the  doubly  homographic  system. 

Then  {ABCD}={A'B'CD'}, 


,  BD.A-D-__BC.A-C_j» 

""^^^"^  AD.B'D'-ACTFC"?      ^' 

Now  taking  O,  an  arbitrary  point  on  the  axis,  let 

OD=ji',  OV>'  =  x',  OK=a,  OA'=a',  0B  =  ^,  0B'=^'. 
ThenBD=x-<^,  B'D'=x'-d',  AD=x-a,  A'D'  =  x'-a', 
(x-d)(x'-a')_p 
{x-a){x'-b')—q' 
which  reduces  to  the  form 

xx'  +  P;r+  Q;r'  +  R =0. 
When  D  and  D'  become  coincident  x'  becomes  equal  to  x 
and  we  have  a  quadratic  from  which  to  determine  x^  i.e.,  the 
positions  of  D  and  D'  when  uniting  to  form  a  double  point. 

Hence  every  doubly  homographic  system  has  two  double 
points  which  are  both  real  or  both  imaginary,  and  of  which 
both  may  be  finite,  or  one  or  both  may  be  at  infinity. 

Evidently  there  cannot  be  more  than  two  double  points, 
for  since  such  points  belong  to  two  systems,  three  double 
points  would  require  the  coincidence  of  three  pairs  of  corre- 
spondents, and  hence  of  all.  (306°) 


HOMOGRAPHY  AND   INVOLUTION.  285 

329°.  If  D  be  one  of  the  double  points  of  a  doubly  homo- 

,.         ,  DB.DA'     CB.C'A'     P 

graphic  system,     jj^-^jj,  =  -,^^^  =  ^,  say. 

Now  DB.DA'  and  DA.DB'  are  respectively  equal  to  the 
squares  on  tangents  from  D  to  any  circles  passing  through 
B,  A'  and  B',  A. 

But  the  locus  of  a  point  from  which  tangents  to  two  given 
circles  are  in  a  constant  ratio  is  a  circle  co-axal  with  both. 

(275°,  Cor.  5> 

Hence  the  following  construction  for  finding  the  double 
points. 

Through  A,  B'  and 
A',  B  draw  any  two 
circles  so  as  to  intersect 
in  two  points  U  and 
V,  and  through  these 
points  of  intersection  pass  the  circle  S",  so  as  to  be  the  locus 
of  a  point  from  which  tangents  to  the  circles  S  and  S'  are  in 
the  given  ratio  VP  :  VQ. 

The  circle  S"  cuts  the  axis  in  D,  D,  which  are  the  required 
double  points. 

Evidently,  instead  of  A,  B'  and  A',  B  we  may  take  any  pairs 
of  non-corresponding  points,  as  A,  C  and  A',  C;  or  B,  C  and 
B',  C.  The  given  ratio  ^/P  :  JQ  is  different,  however,  for 
each  different  grouping  of  the  points. 

Cor.  I.  When  P-Q,  i.e.,  when  ^=^^,   the  circle   S" 

BC     B  (_/ 

takes  its  limiting  form  of  a  line  and  cuts  the  axis  at  one  finite 

point  or  at  none. 

In  this  case  both  double  points  may  be  at  00  or  only  one 

of  them. 

Cor.  2.  If  any  disposition  of  the  constituents  of  the  system 
causes  the  circle  S"  to  lie  wholly  upon  one  side  of  the 
axis,  the  double  points  for  that  disposition  become  imagin- 
ary. 


2^6 


SYNTHETIC   GEOMETRY. 


330°.  Let  L  be  the  axis  of  a  doubly  homographic  system. 

Through  any  point  O  on  the 
circle  S  transfer  the  system, 
by  rectilinear  projection,  to 
the  circle.  Then  ABC..., 
A'B'C...  form  a  doubly 
homographic  system  on  the 
circle. 

Now,  by  connecting  any 
two  pairs  of  non-correspond- 
ents A,  B'  and  A',  B  ;  B,  C 
and  B',  C  ;  C,  A'  and  C,  A, 
we  obtain  the  pascal  line 
KH  which  cuts  the  circle  in 
two  points  such  that 
{KABC}  =  {KA'B'C'}. 

(319°,  Cor.  2) 

Hence  H  and  K  are  double  points  to  the  system  on  the 
circle.  And  by  transferring  K  and  H  back  through  the  point 
O  to  the  axis  L,  we  obtain  the  double  points  D,  D  of  the 
doubly  homographic  range. 

Cor.  I.  When  the  pascal  falls  without  the  circle,  the  double 
points  are  imaginary. 


Cor.  2.  When  one  of  the  joins,  KO  or  HO,  is 
the  double  points  is  at  co . 


to  L,  one  of 


Cor.  3.  If  the  system  upon  the  circle  with  its  double  points 
H  and  K  be  projected  rectilinearly  through  any  point  on 
the  circle  upon  any  axis  M,  it  is  evident  that  the  projected 
system  is  a  doubly  homographic  one  with  its  double  points. 

Cor,  4.  Cor.  3  suggests  a  convenient  method  of  finding  the 
double  points  of  a  given  axial  system. 

Instead  of  employing  a  circle  lying  without  the  axis,  employ 
the  axis  as  a  centre-line  and  pass  the  circle  through  any  pair 
of  non-correspondents. 


HOMOGRAPHY  AND   INVOLUTION.  2Zj 

Then  from  any  convenient  point  on  the  circle  transfer  the 
remaining  points,  find  the  pascal,  and  proceed  as  before. 

331°.  The  following  are  examples  of  the  application  of  the 
double  points  of  doubly  homographic  systems  to  the  solution 
of  problems. 

Ex.  I.  Given  two  non-parallel  lines,  a  point,  and  a  third 
line.  To  place  between  the  non-parallels  a  segment  which 
shall  subtend  a  given  angle  at  the  given  point,  and  be  parallel 
to  the  third  line. 

Let  L  and  M  be  the  non- 
parallel  lines,  and  let  N  be 
the  third  line,  and  O  be  the 
given  point. 

We  are  to  place  a  segment 
between  L  and  M,  so  as  to 
subtend  a  given  angle  at  O  and  be  ||  to  N. 

On  L  take  any  three  points  A,  B,  C,  and  join  OA,  OB,  OC. 
Draw  A«,  B/5,  Qc  all  H  to  N,  and  draw  Oa\  Ob',  Oc'  so  as  to 
make  the  angles  AO^',  B0<^',  COc'  each  equal  to  the  given 
angle. 

Now,  if  with  this  construction  a  coincided  with  <7;',  or  b  with 
//,  or  c  with  c',  the  problem  would  be  solved. 

But,  if  we  take  a  fourth  point  D,  we  have 

OJABCD}  =  {K^CT^')={abcd)  =  {a!b'dd']. 
.'.  abed  and  a' b'c'd'  are  two  homographic  systems  upon  the 
same  axis.     Hence  the  double  points  of  the  system  give  the 
solutions  required. 

Ex.  2.  Within  a  given  A  to  inscribe  a  A  whose  sides  shall 
be  parallel  to  three  given  lines. 

Ex.  3.  Within  a  given  A  to  inscribe  a  A  whose  sides  may 
pass  through  three  given  points. 

Ex.  4.  To  describe  a  A  such  that  its  sides  shall  pass 
through  three  given  points  and  its  vertices  lie  upon  three 
given  lines. 


288  SYNTHETIC   GEOMETRY. 


SYSTEMS  IN  INVOLUTION. 

332°.  If  A,  A',  B,  B'  are  four  points  on  a  common  axis, 
whereof  A  and  A',  as  also  B  and  B',  are  correspondents,  a 
point  O  can  always  be  found  upon  the  axis  such  that 
OA.OA'  =  OB.OB'. 

This  point  O  is  evidently  the  centre  of  the  circle  to  which 
A  and  A',  and  also  B  and  B',  are  pairs  of  inverse  points,  and 
is  consequently  found  by  257°.  "Jp^Vb^ 

Now,  let  P,  P'  be  a  pair  of  variable  conjugate  points  which 
so  move  as  to  preserve  the  relation 

OP.OP'  =  OA.OA'  =  OB.OB'. 
Then  P  and  P'  by  their  varying  positions  on  the  axis  deter- 
mine a   double   system   of  points  C,  C,  D,  D',  E,  E',  etc., 
conjugates  in  pairs,  so  that 

OA .  OA'=OC.  OC'-OD  .  OD'  =  OE .  OE'  =  etc. 

Such  a  system  of  points  is  said  to  be  in  invohition^  and  O 
is  called  the  centre  of  the  involution. 

When  both  constituents  of  any  one  conjugate  pair  lie  upon 
the  same  side  of  the  centre,  the  two  constituents  of  every 
conjugate  pair  lie  upon  the  same  side  of  the  centre,  since 
the  product  must  have  the  same  sign  in  every  case. 

With  such  a  disposition  of  the  points  the  circle  to  which 
conjugatesare  inverse  points  is  real  and  cuts  the  axis  in  two 
_< ,      ,    I      ,        , ,    points  F  and  F'. 

able  conjugates  meet  and  become  coincident 

Hence  the  points  F.  F'  are  the  double  points  or  foci  of  the 
system. 

From  Art.  311°,  i,  FF'  is  divided  harmonically  by  every 
pair  of  conjugate  points,  so  that 

FAF'A',  FBF'B',  etc.,  are  all  harmonic  ranges. 

When  the  constituents  of  any  pair  of  conjugate  points  lie 
upon  opposite  sides  of  the  centre,  the  foci  are  imaginary. 


HOMOGRAPHY  AND   INVOLUTION. 


289 


333°.  Let  A,  A',  B,  B',  C,  C  be  six  points  in  involution,  and 
let  O  be  the  centre. 

Draw  any  line  OPQ  through  O, 
and  take  P  and  O  so  that 

OPrOQ=OA.OA', 
and  join  PA,  PB,  PC,  and  PC, 
and  also  QA',  OB',  QC,  and  QC. 

Then,  •.•    OA.  OA'  =  OP.  0(),    o 

.-.  A,  P,  Q,A'  are  coney clic.    .'.  i.OPA  =  ^OA'Q. 
Similarly,  B,  P,  (2,  B'  are  concycUc,  and  ^OPB  =  ^OB'0,  etc. 

z.APB  =  /.A'OB'. 
Similarly,  ^BPC  =  ^B'QC',  ^CPC'  =  z.C'OC,  etc 

Hence  the  pencils  P(ABCC')  and  (^(A'B'C'C)  are  equianhar- 
monic,  or         {ABCC}  =  {A'B'G'C[. 

Hence  also  {ABB'C}  =  {A'B'BC'J ,  {AA'BCj  =  {A'AB'C'J. 
And  any  one  of  these  relations  expresses  the  condition  that 
the  six  points  symbolized  may  be  in  involution. 

334°.  As  involution  is  only  a  species  of  homography,  the 
relations  constantly  existing  between  homographic  ranges  and 
their  corresponding  pencils,  hold  also  for  ranges  and  pencils 
in  involution.      Hence 

1.  Every  range  in  involution  determines  a  pencil  in  involu- 
tion at  every  vertex,  and  conversely. 

2.  If  a  range  in  involution  be  projected  rectilinearly  through 
any  point  on  a  circle  it  determines  a  system  in  involution  on 
the  circle,  and  conversely. 

Ex.  The  three  pairs  of  opposite  connectors  of  any  four 
points  cut  any  line  in  a  six-point  involution. 

A,  B,  C,  D  are  the  four  points,  v.  p       d  c 

and  P,  F  the  line  cut  by  the  six 
connectors  CD,  DA,  AC,  CB,  BD, 
and  AB.     Then 

D{PQRR'}  =  D{CARB} 
=  B{CARD} 


=  B{0'P'RR'}  =  {P'Q'R'R}, 

T 


(302°) 


290  SYNTHETIC   GKOxMETRY. 

{P()RR'}  =  {P'0'R'R}, 
and  the  six  points  are  in  involution. 

Cor.  I.  The  centre  O  of  the  involution  is  the  radical  centre 
of  any  three  circles  through  PP',  QQ',  and  RR';  and  the 
three  circles  on  the  three  segments  PP',  OQ',  and  RR'  as 
diameters  are  co-axal. 

When  the  order  of  PQR  is  opposite  that  of  P'Q'R'  as  in  the 
figure,  and  the  centre  O  lies  outside  the  points,  the  co-axal 
circles  are  of  the  /./.-species,  and  when  the  two  triads  of 
points  have  the  same  order,  the  co-axal  circles  are  of  the 
f./.-species. 

Cor.  2.  Considering  ABC  as  a  triangle  and  AD,  BD,  CD 
three  lines  through  its  vertices  at  D,  we  have — 

The  three  sides  of  any  triangle  and  three  concurrent  lines 
through  the  vertices  cut  any  transversal  in  a  six-point 
involution. 

Exercises. 

1.  A  circle  and  an  inscribed  quadrangle  cut  any  line  through 

them  in  involution. 

2.  The  circles  of  a  co-axal  system  cut  any  line  through  them 

in  involution. 

3.  Any  three  concurrent  chords  intersect  the  circle  in  six 

points  forming  a  system  in  involution. 

4.  The  circles  of  a  co-axal  system  cut  any  other  circle  in 
.      involution. 

5.  Any  four  circles  through  a  common  point  have  their  six 

radical  axes  forming  a  pencil  in  involution. 


INDEX  OF  DEFINITIONS,  TERMS,  ETC. 


The  Numbers  refer  to  the  Articles. 


Addition  Theorem  for  Sine 

and  Cosine,  . 
Altitude,  . 
Ambiguous  Case, 
Angle,      . 

,,      Acute,  . 

,,      Adjacent  internal 

,,      Basal,    . 

,,       External, 

,,       Obtuse, 

,,       Re-entrant,    . 

„      Right,   . 

,,      Straight, 

,,      Arms  of, 

,,      Bisectors  of,  . 

,,       Complement  of, 

,,      Cosine  of, 

,,      Measure  of,    .        41 

,,      Sine  of, 

,,      Supplement  of, 

,,      Tangent  of,    . 

,,      Vertex  of, 
Angles,  Adjacent,     . 

,,         Alternate,    . 

,,         Interadjacent, 
-    ,,         Opposite,     . 


236 

87 
66 

31 

40 

49 
49 
49 
40 
89 
36 
36 
32 
43 
40 
213 
207 
213 
40 
213 
32 
35 
11 
73 
39 


Angles,  Vertical,  .  .  39 
, ,  Sum  and  Difference  of,  35 
,,       of  the  same  Affection,  65 


Anharmonic  Ratio,  . 

298 

Antihomologous, 

289 

Apothem, 

146 

Arc,          .         .         . 

lOI 

Area, 

136 

,,     of  a  Triangle,. 

1754 

Axiom,     .         .         .         . 

3 

Axis  of  a  Range, 

230 

,,        Perspective, . 

254 

,,        Similitude,   . 

294 

,,        Symmetry,    . 

lOI 

Basal  Angles,  . 

49 

Biliteral  Notation,    . 

22 

Bisectors  of  an  Angle, 

43 

Brianchon's  Theorem, 

320 

Brianchon  Point, 

320 

Centre  of  a  Circle,    . 

92 

,,         Inversion, 

256 

,,         Mean  Position, 

238 

,,         Perspective, 

254 

,,         Simihtude, 

.     281 

291 


292                       SYNTHETIC 

GEOMETRY. 

Centre-line, 

•       95 

Diameter, 

•       95 

Centre-locus,    . 

.     129 

Difference  of  Segments,    .       29 

Centroid, 

•       85 

Dimension, 

•       27 

Circle, 

.       92 

Double  Point, . 

109,  328 

Circle  of  Antisimilitude, 

.     290 

,,        Inversion,  . 

.     256 

Eidograph, 

.  211^ 

,,        Similitude, 

.     288 

End-points, 

22 

Circumangle,    . 

.       36 

Envelope, 

.     223 

Circumcentre,  . 

86,  97 

Equal,      . 

27,  136 

Circumcircle,    . 

.       97 

Equilateral  Triangle 

•       53 

Circumference, 

.       92 

Excircle, . 

.     131 

Circumradius,  . 

•       97 

External  Angle, 

.      49 

Circumscribed  Figure, 

•       97 

Extreme  and  Mean  ] 

Ratio,     183 

Chord,     . 

95 

Extremes, 

•     193 

Chord  of  Contact,    . 

114 

Co-axal  Circles, 

273 

Finite  Line, 

21 

Collinear  Points,       .      13 

I,  247 

Finite  Point,    . 

21 

Commensurable, 

.     150 

Complement  of  an  Angle, 

40 

Generating  Point, 

.       69 

Concentric  Circles,  . 

•       93 

Geometric  Mean, 

.     169 

Conclusion, 

4 

Given  Point  and  Lin 

e,       .       20 

Concurrent  Lines,     .        S 

5,  247 

Congruent, 

51 

Harmonic  Division, 

.     208 

Concyclic, 

97 

Harmonic  Ratio, 

•     299 

Conjugate  Points,  etc.. 

267 

Harmonic  Systems,  . 

313,  314 

Contact  of  Circles,    . 

291 

Homogeneity,  . 

.      160 

Continuity,  Principle  of, 

104 

nomographic  Syster 

IS,     .     327 

Corollary, 

8 

Homologous  Sides,  . 

.     196 

C.-P.  Circles,  . 

274 

Homologous     Lines 

and 

Cosine  of  an  Angle, 

213 

Points, 

.      196 

Constructive  Geometry, 

117 

Hypothenuse,  . 

88,  168 

Curve,      . 

15 

Hypothesis, 

4 

Datum  Line,    . 

20 

Incircle  of  a  Triangle,       .      131 

Degree,    . 

41 

Incommensurable,    . 

.     150 

Desargue's  Theorem, 

253 

Infinity,  . 

21,  220 

Diagonal,          .         .        8 

0,  247 

Initial  Line,     . 

20 

Diagonal  Scale, 

211^ 

Inscribed  Figure, 

•      97 

INDEX   OF   DEFINITIONS,  TERMS,  ETC.        293 


Interadjacent  Angles, 
Inverse  Points,         .      i79> 
Inverse  Figures, 
Involution, 
Isosceles  Triangle,    . 

Join, 

Limit, 

Limiting  Points, 

Line, 

Line  in  Opposite  Senses, 

Line-segment,  . 

Locus, 

L.-P.  Circles,  . 

Magnitude, 

Major  and  Minor,     . 

Maximum, 

Mean  Centre,  . 

Mean  Proportional,  . 

Means,     . 

Measure, 

Median,   . 

Median  Section, 

Metrical  Geometry, 

Minimum, 


73 
256 
260 

332  i 
53  I 

167 

148 

274 
12 

156 
21 
69 

274 

190 
102 
175 
238 
169 
193 
150 
55 
183 
150 
175 


Orthogonal  Projection,  167,  229 


ril6,  Ex.  6 
Nine-points  Circle,  I265,  Ex.  2 

Normal  Quadrangle,         .       89 


Obtuse  Angle, . 
Opposite  Angles, 
Opposite  Internal  Angles, 
Origin,     .         .         .         . 
Orthocentre,     . 
Orthogonally,  . 


40 
39 
49 
20 
88 
115 


Pantagraph,      . 
Parallel  Lines, 
Parallelogram, 
Pascal's  Hexagram 
Pascal  Line,     . 
Peaucellier's  Cell, 
Pencil, 
Perigon,  . 
Perimeter, 
Perspective, 
Perspective,  Axis  of, 

,,  Centre  of. 

Perpendicular, . 
Physical  line,  . 
Plane,       .         •         •  10, 

Plane  Geometric  Figure 
Plane  Geometry, 
Point, 

Point  of  Bisection, 
,,       Contact, 

Point,  Double,  .      109 

Pole, 

Polar, 

Polar  Reciprocal 

Polar  Circle,  Centre,  etc. 

Polygon, . 

Prime  Vector,  . 

Projection, 

Proportion, 

Proportional  Compasses 

Protractor, 


70 

80 
319 
319 
211 
203 

36 
146 
254 
254 
254 

40 

12 

7 

10 

II 

13 

30 

109 

328 

266 

266 

268 

266 

132 

20 

167 
192 

2Ili 
123 


—r 


Quadrangle,  ^ 
Quadrilateral,  j 
Quadrilateral,  complete, 


80,  89 
.     247 


!94 


SYNTHETIC   GEOMETRY. 


Uadian,    . 

Radical  Axis,   . 

Radical  Centre. 

Radius,    . 

Radius  Vector, 

Range,     . 

Ratio, 

Reciprocation, 

Reciprocal  Segments 

Rectangle, 

Rectilinear  Figure, 

Rediidio  ad  abstu-dtuii , 

Re-entrant  Angle, 

Rhombus, 

Right  Angle,    . 

Right  Bisector, 

Rotation  of  a  Line, 

Rule, 

Rule  of  Identity, 


Scalene  Triangle 
Secant,     . 
Sector,     . 
Segment, 
Semicircle, 
Self-conjugate 
Self-reciprocal 
Sense  of  a  Line, 
Sense  of  a  RectangI 
Similar  Figures, 
Similar  Triangles, 
Sine  of  an  Angle, 
Spatial  Figure, 


"'1 

alj 


Square, 


78, 


207 

273 

276 

92 

32 

230 

188 

321 

184 

82 

14 

54 

89 

82 

36 

42 

2,  222 

16 

7 

57 

95 

211^ 

21 

102 


268 

156 
161 
206 

n.  196 
213 

19 
82, 121 


Straight  Angle,         .         .  36 
„       Line,  .         .         .14 

Edge,          .         .  16 

Sum  of  Line-segments,      .  28 

Sum  of  Angles,         .         .  35 

Superposition, ...  26 

Supplement  of  an  Angle,  40 

Surface,    ....  9 

Tangent,           .         .         .  109 

Tangent  of  an  Angle,        .  213 

Tensor,    .         .         .         .189 

Tetragram,  —  .         .         .  247 

Theorem,          ...  2 

Trammel,  .         .         .107 

'Transversal,     •         •         •  73 

Trapezoid,         ...  84 

Triangle,           ...  48 

,,         Acute-angled,  etc.,  77 

,,         Base  of,    .         .  49 

,,         Equilateral,       .  53 

,,         Isosceles,.         .  53 

,,         Obtuse-angled,  TJ 

Scalene,    .         .  57 

,,         Right- Angled,  .  77 

Uniliteral  Notation,          .  li 
Unit-area,     \           ^ 
Unit-length,  / 

Versed  Sine  of  an  Arc,      .  176 

Vertex  of  an  Angle, .         .  32 

Vertex  of  a  Triangle,         .  49 

Vertical  Angles,        .         .  39 


PRINTED   BY    ROBERT   MACLEHOSE,    UNIVERSITY   PRESS,  GLASGOW. 


May  189: 


A  Catalogue 

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CONTENTS 


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Elementary  Classics  . 
Classical  Series  ,        .        .        . 
Classical  Library  ;  Texts,  Com- 
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Italian 

Spanish 

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Algebra        

Euclid  and  Pure  Geometry 
Geometrical  Drawing 
Mensuration         .        .        .        . 
Trigonometry       .       .        .        , 
Analytical  Geometry 
Problems  and  Questions  in  Ma- 
thematics   

Higher  Pure  Mathematics 
Mechanics 


page 
2 
4 


12 


Physics  . 

Astronomy 

Historical 


32 


NATURAL  Sciences- 
Chemistry     .        .  •     ,        .        .83 
Physical  Geography,  Geology, 
AND  Mineralogy       ...      85 

Biology 36 

Medicine 38 

HUMAN  sciences- 
Mental  AND  Moral  Philosophv  39 
Political  Economy  ...  41 
Law  and  Politics  ...  42 
Anthropology  ....  48 
Education 48 

Technical  Knowledge- 
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ing        

Military  and  Naval  Science 
Agriculture 
Domestic  Economy 
Book-Keeping 


Geography 

HISTORY  . 

Art  . 
divinity 


CLASSICS. 

Elementary  Classics ;  Classical  Series ;  Classical  Library,  (1)  Texts,  (2)  Trans- 
lations; Grammar,  Composition,  and  PMlology;  Antiquities,  Ancient 
History,  and  Philosophy. 

^ELEMENTARY    CLASSICS. 

18mo,  Eighteenpence  each. 

The  following  contain  Introductions,  Notes,  and  Vocabularies,  and 
in  some  cases  Exercises. 

ACCIDENCE,  LATIN,  AND  EXERCISES  ARRANGED  FOR  BEGINNERS.— By 

W.  Welch,  M.A.,  and  C.  G.  Duffield,  M.A. 
AESCHYLUS.— PROMETHEUS  VINCTUS.     By  Rev.  H.  M.  Stephenson,  M.A. 
ARRIAN.— SELECTIONS.     With  Exercises.    By  Rev.  John  Bond,  M.A.,  and 

Rev.  A.  S.  Walpole,  M.A. 
AULUS  GELLIUS,  STORIES  FROM.— Adapted  for  Beginners.     With  Exercises. 

By  Rev.  G.  H.  Nall,  M.A.,  Assistant  Master  at  Westminster. 
CiESAR.— THE   HELVETIAN  WAR.    Being  Selections  from  Book  L  of  The 
Gallic  War.    Adapted  for  Beginners.    With  Exercises.     By  W.  Welch,  M.A., 
and  0.  G.  Duffield,  M.A. 
THE  INVASION  OF  BRITAIN.    Being  Selections  from  Books  IV.  and  V.  of  The 
Gallic  War.    Adapted  for  Beginners.     With  Exercises.     By  W.  Welch,  M.A., 
and  C.  G.  Duffield,  M.A. 
SCENES  FROM  BOOKS  V.  and  VI.     By  0.  Colbeck,  M.A. 
THE  GALLIC  WAR.     BOOK  I.     By  Rev.  A.  S.  Walpole,  M.A. 
BOOKS  II.  AND  IIL     By  the  Rev.  W.  G.  Rutherford,  M.A.,  LL.D. 
BOOK  IV.     By  Clement  Bryans,  M.A.,  Assistant  Master  at  Dulwich  College. 
BOOK  V.     By  C.  Colbeck,  M.A.,  Assistant  Master  at  Harrow. 
BOOK  VI.     By  the  same  Editor. 

BOOK  VIL     By  Rev.  J.  Bond,  M.A.,  and  Rev.  A.  S.  Walpole,  M.A. 
THE  CIVIL  WAR.     BOOK  L     By  M.  Montgomery,  M.A.         [In  preparation. 
CICERO.— DE  SENECTUTE.     By  E.  S.  Shuckburgh,  M.A. 
DE  AMICITIA.     By  the  same  Editor. 

STORIES  OF  ROMAN  HISTORY.     Adapted  for  Beginners.    With  Exercises. 
By  Rev.  G.  B.  Jeans,  M.A.,  and  A.  V.  Jones,  M.A. 
EURIPIDES.— ALCESTIS.     By  Rev.  M.  A.  Bayfield,  M.A. 
MEDEA.    By  A.  W.  Verrall,  Litt.D.,  and  Rev.  M.  A.  Bayfield,  M.A. 

[In  the  Press. 
HECUBA.     By  Rev.  J.  Bond,  M.A,,  and  Rev.  A.  S.  Walpole,  M.A. 
EUTROPIUS.— Adapted  for  Beginners.     With  Exercises.     By  W.  Welch,  M.A., 

and  C.  G.  Duffield,  M.A. 

HERODOTUS.      TALES  FROM  HERODOTUS.     Atticised  by  G.  S.  Farnell, 

M.A.  [In  the  Press. 

HOMER.— ILIAD.   BOOK  I.   By  Rev.  J.  Bond,  M. A.,  and  Rev.  A.  S.  Walpole,  M.A. 

BOOK  XVIII.    By  S.  R.  James,  M.A.,  Assistant  Master  at  Eton. 

ODYSSEY.    BOOK  I.    By  Rev.  J.  Bond,  M.A.,  and  Rev.  A.  S.  Walpole,  M.A. 


ELEMENTARY  CLASSICS  8 

HORACE.— ODES.     BOOKS  I. -IV.     By  T.  E.  Page,  M.A.,  Assistant  Master 

at  the  Charterhouse.     Each  Is.  6d. 
LIVY.— BOOK  I.     By  H.  M.  Stephenson,  M.A. 
BOOK  XXI.    Adapted  from  Mr.  Capes's  Edition.    By  J.  E.  Melhuish,  M.A. 
BOOK  XXII.     By  the  same. 
THE  HANNIBALIAN  WAR.     Being  part  of  the  XXI.  and  XXII.  BOOKS  OF 

LIVY  adapted  for  Beginners.     By  G.  C.  Macaxjlay,  M.A. 
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LIVY,  adapted  for  Beginners.     With  Exercises.     By  G.  Richards,  M.A.,  and 

Rev.  A.  S.  Walpole,  M.A. 
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By  H.  Wilkinson,  M.A. 
LUCIAN.— EXTRACTS  FROM  LUCIAN.  With  Exercises.  By  Rev.  J.  Bond,  M.A., 

and  Rev.  A.  S.  Walpole,  M.A. 
NEPOS.— SELECTIONS  ILLUSTRATIVE  OF  GREEK  AND  ROMAN  HISTORY. 

With  Exercises.     By  G.  S.  Farnell,  M.A. 
OVID.— SELECTIONS.     By  E.  S.  Shuckburgh,  M.A. 
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M.A.,  and  Rev.  A.  S.  Walpole,iM.A. 
PH.ffiDRUS.  —  SELECT  FABLES. '  Adapted   for    Beginners.     With    BxercLses. 

By  Rev.  A.  S.  Walpole,  M.A. 
THUCYDIDES.— THE  RISE  OF  THE  ATHENIAN  EMPIRE.     BOOK  I.     Chs. 

89-117  and  228-238.    With  Exercises.     By  F.  H.  Colson,  M.A. 
VIRGIL.— SELECTIONS.     By  E.  S.  Shuckburgh,  M.A. 
BUCOLICS.    By  T.  E.  Page,  M.A. 
GEORGICS.    BOOK  L    By  the  same  Editor. 
BOOK  II.     By  Rev.  J.  H.  Skeine,  M.A. 
iENEID.     BOOK  I.    By  Rev.  A.  S.  Walpole,  M.A. 
BOOK  II.     By  T.  E.  Page,  M.A. 
BOOK  III.     By  the  same  Editor. 
BOOK  IV.     By  Rev.  H.  M.  Stephenson,  M.A. 
BOOK  V.    By  Rev.  A.  Calvert,  M.A. 
BOOK  VI.    By  T.  E.  Page,  M.A. 
BOOK  VII.     By  Rev.  A.  Calvert,  M.A. 
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BOOK  IX.    By  Rev.  H.  M.  Stephenson,  M.A. 
BOOK  X.     By  S.  G.  Owen,  M.A. 
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By  W.  Welch,  M.A.,  and  0.  G.  Duffield,  M.A. 
BOOK  I.     With  Exercises.     By  E.  A.  Wells,  M.A. 
BOOK  L     By  Rev.  A.  S.  Walpole,  M.A. 
BOOK  II.     By  the  same  Editor. 
BOOK  III.    By  Rev.  G.  H.  Nall,  M.A. 
BOOK  IV.     By  Rev.  E.  D.  Stone,  M.A. 

SELECTIONS  FROM  BOOK  IV.    With  Exercises.     By  the  same  Editor. 
SELECTIONS  FROM  THE  CYROPiEDIA.    With  Exercises.     By  A.  H.  Cooke, 

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The  following  contain  Introductions  and  Notes,  but  no  Vocabu- 
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HERODOTUS.— SELECTIONS  FROM  BOOKS  VIL  and  VIII.    THE  EXPEDI- 
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HORACE.— SELECTIONS  FROM  THE  SATIRES  AND  EPISTLES.     By  Rev.  W. 
J.  V.  Baker,  M.A. 
SELECT  EPODBS  AND  ARS  POETICA.    By  H.  A.  Dalton,  M.A.,  Assistant 
Master  at  Winchester. 


4  CLASSICS 

PLATO.— EUTHYPHRO  AND  MENEXENUS.     By  C.  E.  Graves,  M.A. 
TERENCE. —SCENES  FROM  THE  ANDRIA,     By  F.  W.  Cornish,  M.  A. ,  Assistant 

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THE    GREEK   ELEGIAC    POETS.— FROM  CALLINUS    TO    CALLIMACHUS. 

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THUCTDroES.— BOOK  IV.  Chs.  1-41.    THE  CAPTURE  OF  SPHACTERIA.    By 

C.  E.  Graves,  M.A. 

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6  CLASSICS 

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CLASSICAL   LIBRARY. 

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CLASSICAL  LIBRARY  7 

THE  ETHICS.    Translated  by  Rev.  J.  E.  C.  Welldon,  M.A.    Cr.  8vo.  [In  prep, 
THE  SOPHISTICI  ELENCHI.     With  Translation.    By  E.  Postb,  M.A.,  Fellow 

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ON  THE  CONSTITUTION  OP  ATHENS.    Edited  by  J.  B.  Sandys,  Litt.D. 
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MELEAGER.— FIFTY  POEMS  OF  MELEAGER.     Translated  by  Walter  Head- 
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8  CLASSICS 

PAUSANIAS.— DESCRIPTION    OF   GREECE.      Translated  with  Commentary 
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Ex.  fcap.  8vo.     5s. 
KEY,  for  Teachers  only.     Ex.  fcap.  Svo.    4s.  6d. 
LUPTON.— *AN   INTRODUCTION   TO   LATIN    ELEGIAC  VERSE  COMPOSI- 
TION.    By  J.  H.  LuPTON,  Sur-Master  of  St.  Paul's  School.     Gl.  Svo.    2s.  6d. 
KEY  TO  PART  II.  (XXV. -C.)    Gl.  Svo.     3s.  6d. 
*AN  INTRODUCTION  TO  LATIN  LYRIC  VERSE  COMPOSITION.     By  the 

same.     Gl.  Svo.     3s.     KEY,  for  Teachers  only.    Gl.  Svo.     4s.  6d. 
MAOKIE.— PARALLEL    PASSAGES    FOR    TRANSLATION    INTO    GREEK 
AND  ENGLISH.    "With  Indexes.     By  Rev.  Ellis  C.  Mackie,  M.A.,  Classical 
Master  at  Heversham  Grammar  School.     Gl.  Svo.    4s.  6d. 
*MAOMILLAN.— FIRST    LATIN    GRAMMAR.      By  M.   C.    Maomillan,    M.A 

Fcap.  Svo.     Is.  6d. 
MACMILLAN'S  GREEK  COURSE.— Edited  by  Rev.  W.  G.  Rutherford,  M.A., 
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*FIRST  GREEK  GRAMMAR— ACCIDENCE.    By  the  Editor.    23. 
*FIRST  GREEK  GRAMMAR— SYNTAX.     By  the  same.     2s. 
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*EASY  EXERCISES  IN  GREEK  ACCIDENCE.     By  H.  G.  Underbill,  M.A., 

Assistant  Master  at  St.  Paul's  Preparatory  School.     2s. 
*A  SECOND   GREEK   EXERCISE    BOOK.     By    Rev.   W.  A.    Heard,    M.A., 

Headmaster  of  Fettes  College,  Edinburgh.     2s.  6d. 
EASY  EXERCISES  IN  GREEK   SYNTAX.      By  Rev.   G.   H.   Nall,   M.A., 
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MANUAL  OF  GREEK  ACCIDENCE.     By  the  Editor.  [In  preparation. 

MANUAL  OF  GREEK  SYNTAX.     By  the  Editor.  [In  preparation. 

ELEMENTARY  GREEK  COMPOSITION.    By  the  Editor.         [In  preparation. 
'^MACMILLAN'S  GREEK  READER.— STORIES  AND  LEGENDS.    A  First  Greek 
Reader,  with  Notes,  Vocabulary,  and  Exercises.     By  F.  H.  Colson,    M.A., 
Headmaster  of  Plymouth  College.    Gl.  Svo.    3s. 
IVLACMILLAN'S  LATIN  COURSE.— By  A.  M.  Cook,  M.A.,  Assistant  Master  at 
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*FIRST  PART.     Gl.  Svo.     3s.  6d. 
♦SECOND  PART.    2s.  6d.  [Third  Part  in  preparation. 

*MAOMILLAN'S  SHORTER  LATIN  COURSE.— By  A.  M.  Cook,  M.A.  Being  an 
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*MAC]yiILLAN'S  LATIN  READER.— A  LATIN  READER  FOR  THE  LOWER 
FORMS  IN  SCHOOLS.  By  H.  J.  Hardy,  M.A.,  Assistant  Master  at  Win- 
chester.    Gl.  Svo.     2s.  6d, 

*MARSHALL.— ATABLE  OF  IRREGULAR  GREEK  VERBS,  classified  according 
to  the  arrangement  of  Curtius's  Greek  Grammar.     By  J.  M.  Marshall,  M.A., 
Headmaster  of  the  Grammar  School,  Durham.    Svo.     Is. 
MAYOR.— FIRST  GREEK  READER.    By  Prof.  John  E.  B.  Mayor,  M.A.,  Fellow 
of  St.  John's  College,  Cambridge.     Fcap.  Svo.    4s.  6d. 

MAYOR.— GREEK  FOR  BEGINNERS.  By  Rev.  J.  B.  Mayor,  M.A.,  late 
Professor  of  Classical  Literature  in  King's  College,  London.  Part  I.,  with 
Vocabulary,  Is.  6d.  Parts  II.  and  III.,  with  Vocabulary  and  Index.  Fcap. 
Svo.  3s.  6d.  Complete  in  one  Vol.  4s.  6d. 
NIXON.— PARALLEL  EXTRACTS,  Arranged  for  Translation  into  English  and 
Latin,  with  Notes  on  Idioms.  By  J.  E.  Nixon,  M.A.,  Fellow  and  Classical 
Lecturer,  King's  College,  Cambridge.  Part  I.— Historical  and  Epistolary, 
Cr.  Svo.    3s.  6d. 


GRAMMAR,  COMPOSITION,  AND  PHILOLOGY  11 

PROSE  EXTRACTS,  Arranged  for  Translation  into  English  and  Latin,  with 

General  and  Special  Prefaces  on  Style  and  Idiom.    By  the  same.    I,  Oratorical. 

II.   Historical.     III.   Philosophical.     IV.   Anecdotes  and  Letters.    2d  Ed., 

enlarged  to  280  pp.    Cr.  8vo.    4s.  6d.    SELECTIONS  FROM  THE  SAME.    3s. 
Translations  of  about  70  Extracts  can  be  supplied  to  Schoolmasters  (2s.  6d.), 

on  application  to  the  Author  :  and  about  40  similarly  of  "Parallel  Extracts," 

Is.  ()d.  post  free. 
*PANTIN.— A  FIRST   LATIN  VERSE   BOOK.     By  W.    B.   P.   Pantin,  M.A., 

Assistant  Master  at  St.  Paul's  School.     Gl.  Svo.     Is.  6d. 
*PEILE.— A  PRIMER  OP  PHILOLOGY.     By  J.  Peile,  Litt.D.,  Master  of  Christ's 

College,  Cambridge.     ISmo.     Is. 
*POSTGATE.— SERMO  LATINUS.    A  short  Guide  to  Latin  Prose  Composition. 

By  Prof.  J.  P.  PosTGATE,  Litt.D.,  Fellow  of  Trinity  College,  Cambridge.    Gl. 

Svo.     2s.  6d.     KEY  to  "  Selected  Passages."    Gl.  Svo.     3s.  6d. 
POSTGATE  and  VINCE.— A  DICTIONARY  OF    LATIN    ETYMOLOGY.     By 

J.  P.  PosTQATE  and  C.  A.  Vince.  [In  preparation. 

POTTS.— *HINTS  TOWARDS  LATIN  PROSE  COMPOSITION.    By  A.  W.  Potts, 

M.A.,  LL.D.,  late  Fellow  of  St.  John's  College,  Cambridge.    Ex.  fcap.  Svo.   3s. 
*PASSAGES  FOR  TRANSLATION  INTO  LATIN  PROSE.    Edited  with  Notes 

and  References  to  the  above.    Ex.  fcap.  Svo.    2s.  6d.     KEY,  for  Teachers  only. 

2s.  6d. 
*PRESTON.— EXERCISES  IN  LATIN  VERSE  OF  VARIOUS  KINDS.    By  Rev. 

G.  Preston.    Gl.  Svo.     2s.  6d.     KEY,  for  Teachers  only.    Gl.  Svo.    5s. 
REID.— A  GRAMMAR  OF  TACITUS.     By  J.  S.  Reid,  Litt.D.,  Fellow  of  Caius 

College,  Cambridge.  [In  the  Press. 

A  GRAMMAR  OF  VIRGIL.    By  the  same.  [In  preparation. 

ROBY. — Works  by  H.  J.  Roby,  M.A.,  late  Fellow  of  St.  John's  College,  Cambridge. 

A  GRAMMAR  OF  THE  LATIN  LANGUAGE,  from  Plautus  to  Suetonius.     Part 

I.      Sounds,  Inflexions,  Word -formation,  Appendices.   Cr.  Svo.    9s.    Part  II. 

Syntax,  Prepositions,  etc.     10s.  6d. 
*SCHOOL  LATIN  GRAMMAR.     Cr.  Svo.    5s. 

AN  ELEMENTARY  LATIN  GRAMMAR.  [In  the  Press. 

*RUSH.— SYNTHETIC  LATIN  DELECTUS.    With  Notes  and  Vocabulary.    By  E. 

Rush,  B.A.    Ex.  fcap.  Svo.    2s.  6d. 
*RUST.— FIRST  STEPS  TO  LATIN  PROSE  COMPOSITION.    By  Rev.  G.  Rust, 

M.A.    ISmo.    Is.  6d.    KEY,  for  Teachers  only.    ByW.  M.Yates.    ISmo.    Ss.  6d. 
RUTHERFORD.— Works  by  the  Rev.  W.  G.  Rutherford,  M.A.,  LL.D.,  Head- 

master  of  Westminster. 
REX  LEX.     A  Short  Digest  of  the  principal  Relations  between  the  Latin, 

Greek,  and  Anglo-Saxon  Sounds.    Svo.  [In  preparation. 

THE  NEW  PHRYNICHUS ;  being  a  Revised  Text  of  the  Ecloga  of  the  Gram- 

marian  Phrynichus.    With  Introduction  and  Comnientary.     Svo.    18s.    (See 

also  Macmillan's  Greek  Course.) 
SHUCKBURGH.— PASSAGES  FROM  LATIN  AUTHORS  FOR  TRANSLATION 

INTO  ENGLISH.    Selected  with  a  view  to  the  needs  of  Candidates  for  the 

Cambridge  Local,  and  Public  Schools'  Examinations.     By  E.  S.  Shuckburgh, 

M.A.    Cr.  Svo.    2s. 
*SIMPSON.  — LATIN  PROSE  AFTER  THE  BEST  AUTHORS  :  Caesarian  Prose. 

By  F.  P.  Simpson,  B.A.     Ex.  fcap.  Svo.    2s.  6d,     KEY,  for  Teachers   only. 

Ex.  fcap.  Svo.     5s. 
STRACHAN  and  "WTLKINS.- ANALECTA.    Selected  Passages  for  Translation. 

By  J.  S.  Strachan,  M.A.,  Pi'ofessor  of  Greek,  and  A.  S.  Wilkins,  Litt.D., 

Professor  of  Latin  in  the  Owens  CoUege,  Manchester.    Cr.  Svo.    5s.    KEY  to 

Latin  Passages.     Cr.  Svo.    6d. 
THRING. — Works  by  the  Rev.  E.  Thrino,  M.A.,  late  Headmaster  of  Uppingham. 
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Coloured  Sentence  Maps.    Fcap.  Svo.     2s.  6d. 
A  MANUAL  OF  MOOD  CONSTRUCTIONS.     Fcap.  Svo.     Is.  6d. 
"WELCH    and    DUFPIELD.  —  LATIN   ACCIDENCE    AND    EXERCISES   AR- 

RANGED    FOR    BEGINNERS.      By   W.    Welch    and    C.    G.    Duffield, 

Assistant  Masters  at  Crauleigh  School.     ISmo.     Is.  6d. 


12  CLASSICS 

WHITE.— FIRST  LESSONS  IN  GREEK.  Adapted  to  Goodwin's  Greek  Gram- 
mar, and  designed  as  an  introduction  to  the  Anabasis  of  Xenophon.  By 
John  Williams  White,  Assistant  Professor  of  Greek  in  Harvard  University, 
U.S.A.  Or.  8vo.  3s.  6d. 
WEIGHT.— Works  by  J.  Wright,  M.A,  late  Headmaster  of  Sutton  Coldfield 
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A  HELP  TO  LATIN  GRAMMAR ;  or,  the  Form  and  Use  of  Words  in  Latin, 
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THE  SEVEN  KINGS  OF  ROME.  An  Easy  Narrative,  abridged  from  the  First 
Book  of  Livy  by  the  omission  of  Difficult  Passages  ;  being  a  First  Latin  Read- 
ing Book,  with  Grammatical  Notes  and  Vocabulary.     Fcap.  8vo.     3s.  6d. 

FIRST  LATIN  STEPS;  or,  AN  INTRODUCTION  BY  A  SERIES  OF 
EXAMPLES  TO  THE  STUDY  OF  THE  LATIN  LANGUAGE.    Or.  8vo.    3s. 

ATTIC  PRIMER.     Arranged  for  the  Use  of  Beginners.     Ex.  fcap.  Svo.    2s.  6d. 

A  COMPLETE  LATIN  COURSE,  comprising  Rules  with  Examples,  Exercises, 
both  Latin  and  English,  on  each  Rule,  and  Vocabularies.     Cr.  Svo.     2s.  6d. 

ANTIQUITIES,  ANCIENT  HISTOBY,  AND 
PHILOSOPHY. 

ARNOLD.— A   HANDBOOK    OF  LATIN  EPIGRAPHY.      By  W.   T.   Arnold, 

M.A.  [771  preparation. 

THE    ROMAN   SYSTEM    OF   PROVINCIAL   ADMINISTRATION    TO   THE 

ACCESSION  OF  CONSTANTINE  THE  GREAT.    By  the  same.    Cr.  Svo.     6s. 

ARNOLD.— THE  SECOND  PUNIC  WAR.     Being  Chapters  from  THE  HISTORY 

OF  ROME  by  the   late    Thomas    Arnold,   D.D.,    Headmaster   of  Rugby. 

Edited,  with  Notes,  by  W.  T.  Arnold,  M.A.    With  8  Maps.    Cr.  Svo.    5s. 

*BEESLY.— STORIES  FROM  THE    HISTORY  OF  ROME.     By  Mrs.  Beeslt. 

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BLACKIE.— HOR.E  HELLENICS.     By  John  Stuart  Blackie,  Emeritus  Pro- 
fessor of  Greek  in  the  University  of  Edinburgh.    Svo.    12s. 
BURN.— ROMAN  LITERATURE  IN  RELATION  TO  ROMAN  ART.    By  Rev. 
Robert  Burn,  M.A.,  late  Fellow  of  Trinity  College,  Cambridge.    Illustrated. 
Ex.  cr.  Svo.     14s. 
BURY.— A  HISTORY  OF  THE  LATER  ROMAN  EMPIRE  FROM  ARCADIUS 
TO  IRENE,  A.D.  395-800.     By  J.  B.  Bury,  M.A.,  Fellow  of  Trinity  College, 
Dublin.     2  vols.    Svo.     32s. 
*OLASSICAL  WRITERS.— Edited  by  John  Richard  Green,  M.A.,  LL.D.    Fcap. 
Svo.     Is.  6d.  each. 
SOPHOCLES.     By  Prof.  L.  Campbell,  M.A. 
EURIPIDES.     By  Prof.  Mahaffy,  D.D. 
DEMOSTHENES.     By  Prof.  S.  H.  Butcher,  M.A. 
VIRGIL.     By  Prof.  Nettleship,  M.A. 
LIVY     By  Rev.  W.  W.  Capes,  M.A. 

TACITUS.     By  Prof.  A.  J.  Church,  M.A.,  and  W.  J.  Brodribb,  M.A. 
MILTON.     By  Rev.  Stopford  A  Brooke,  M.A. 
FREEMAN.— Works  by  Edward  A.  Freeman,  D.C.L.,  LL.D.,  Regius  Professor  of 
Modern  History  in  the  University  of  Oxford. 
HISTORY  OF  ROME.     (Historical  Course  for  Schools.)    18mo.     [In  i^reparation. 
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HISTORICAL  ESSAYS.     Second  Series.     [Greek  and  Roman  History.]     Svo. 
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FYFPE.— A  SCHOOL  HISTORY  OF  GREECE.    By  C.  A.  Fyffe,  M.A    Cr.  Svo. 

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GARDNER.— SAMOS  AND  SAMIAN  COINS.    An  Essay.     By  Percy  Gardner, 
Litt.D.,  Professor  of  Archaeology  in  the  University  of  Oxford.    With  Illustra- 
tions.   Svo.     7s.  6d. 


ANCIENT  HISTORY  AND  PHILOSOPHY  13 

G-EDDES.— THE  PROBLEM  OF  THE  HOMERIC  POEMS.     By  W.  D.  Geddks, 

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GOW.— A  COMPANION  TO  SCHOOL  CLASSICS.    By  James  Gow,  Litt.D., 
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JEBB. — Works  by  R.  C.  Jebb,  Litt.D.,  Professor  of  Greek  in  the  University  of 
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ledge.    By  Dr.  Carl  Shuchhardt.     Translated  by  Eugenie  Sellers.     Intro- 
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SHUCKBURGH.— A  SCHOOL  HISTORY  OF  ROME.      By  E.  S.  Shuckburgh, 
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14  MODERN  LANGUAGES  AND  LITERATURE 

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*MAOMILLAN'S  FRENCH  COMPOSITION.— By  G.  E.  Fasnacht.  Ex.  fcap. 
8vo.    Part  I.  Elementary.     2s.  6d.     Part  II.    Advanced.  [In  the  Press. 

THE  TEACHER'S  COMPANION  TO  MACMILLAN'S  COURSE  OF  FRENCH 
COMPOSITION.     By  G.  E.  Fasnacht.    Part  I.     Ex.  fcap.  8vo.    4s.  6d. 
MACMILLAN'S  PROGRESSIVE  FRENCH  READERS.    By  G.  E.  Fasnacht.    Ex. 

fcap.  8vo. 
*FiRST  Year,  containing  Tales,  Historical  Extracts,  Letters,  Dialogues,  Ballads, 
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(2)  in  alphabetical  order.     With  Imitative  Exercises.     2s.  6d. 
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MACMILLAN'S  FOREIGN  SCHOOL  CLASSICS.    Edited  by  G.  B.   Fasnacht. 
18mo. 
*CORNBILLB— LB  CID.    By  G.  B.  Fasnacht.     Is. 

*DUMAS— LBS  DEMOISELLES  DE  ST.  CYR.    By  Victor  Oger,  Lecturer  at 
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Assistant  Master  at  Harrow.  [In  preparation. 

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»FRBNCH  READINGS  FROM  ROMAN  HISTORY.      Selected    from  various 

Authors,  by  C.  Colbeck,  M.A.,  Assistant  Master  at  Harrow.    4s.  Cd. 
»SAND,   GEORGE— LA  MARE  AU  DIABLB.      By    W.    E.    Russell,  M.A., 
Assistant  Master  at  Haileybury.     Is. 


20  MODERN  LANGUAGES  AND  LITERATURE 

*SANDEAU,   JULES— MADEMOISELLE    DE    LA  SEIGLI^RE.      By   H.    C. 

Steel,  Assistant  Master  at  Winchester.     Is.  6d. 
THIBRS'S  HISTORY  OP  THE  EGYPTIAN  EXPEDITION.    By  Rev.   H.  A. 
Bull,  M.  A.,  formerly  Assistant  Master  at  Wellington.  [In  preparation. 

*VOLTAIRE— CHARLES  XII.    By  G.  E.  Fasnacht.    8s.  6d. 
*MASSON.— A  COMPENDIOUS  DICTIONARY  OP  THE  PRENCH  LANGUAGE. 
Adapted  from  tlie  Dictionaries  of  Professor  A.  Elwall,    By  Gustave  Masson. 
Cr.  8vo.     6s. 
MOLIERE.— LE  MALADE  IMAGINAIRE.    With  Introduction  and  Notes,  by  P. 

Tabver,  M.A.,  Assistant  Master  at  Eton.     Fcap.  8vo.    2s.  6d. 
*PELLISSIER.— FRENCH    ROOTS  AND    THEIR    FAMILIES.      A   Synthetic 
Vocabulary,  based  upon  Derivations.      By  E.  Pellissieb,  M.A.,  Assistant 
Master  at  Clifton  College.     Gl.  8vo.     6s. 

GERMAN. 

BEHAGEL.— THE  GERMAN  LANGUAGE.  By  Dr.  Otto  Behagel.  Translated 
by  Emil  Tbechmann,  B.A.,  Ph.D.,  Lecturer  in  Modern  Literature  in  the 
University  of  Sydney,  N.S.  W.    Gl.  Svo.  [In  the  Press. 

HUSS.— A    SYSTEM  OF  ORAL  INSTRUCTION  IN  GERMAN,  by  means  of 
Progressive  Illustrations  and  Applications  of  the  leading  Rules  of  Grammar. 
By  H.  C.  O.  Huss,  Ph.D.    Cr.  Svo.    5s. 
MAOMILLAN'S  PROGRESSIVE  GERMAN  COURSE.    By  G.  E.  Pasnacht.    Ex. 
fcap.  Svo. 
*FiBST  Yeab.    Easy  lessons  and  Rules  on  the  Regular  Accidence.    Is.  6d. 
*Second  Yeab.    Conversational  Lessons  in  Systematic  Accidence  and  Elementary 
Syntax.      With   Philological    Illustrations    and    Etymological    Vocabulary. 
33.  6d. 
Thtbd  1''eab.  [In  the  Press. 

TEACHER'S   COMPANION  TO   MACMILLAN'S   PROGRESSIVE   GERMAN 
COURSE.    With  copious  Notes,  Hints  for  Different  Renderings,  Synonyms, 
Philological  Remarks,  etc.    ByG.  E.  Faskacht.     Ex.  fcap.  Svo.     Fibst  Y'ear. 
4s.  6d.    Second  Yeab.    4s.  6d. 
MACMILLAN'S  GERMAN  COMPOSITION.     By  G.  E.  Pasnacht.    Ex.  fcap.  Svo. 
*I.  FIRST  COURSE.     Parallel  German-English  Extracts  and  Parallel  English- 
German  Syntax.     2s.  6d. 
TEACHER'S  COMPANION  TO  MACMILLAN'S  GERMAN  COMPOSITION. 
By  G.  E.  Pasnacht.    Fibst  Course.    Gl.  Svo.    4s.  6d. 
MACMILLAN'S  PROGRESSIVE  GERMAN  READERS.    ByG.  E.  Pasnacht.   Ex. 
fcap.  Svo. 
*Fibst  Yeab,  containing  an  Introduction  to  the  German  order  of  Words,  with 
Copious  Examples,  extracts  from  German  Authors  in  Prose  and  Poetry;  Notes, 
and  Vocabularies.    2s.  Qd. 
MACMILLAN'S  PRIMARY  SERIES  OF  GERMAN  READING  BOOKS.     Edited 
by  G.  E.  Pasnacht.     With  Notes,  Vocabularies,  and  Exercises.    Gl.  Svo. 
*GRIMM— KINDER  UND  HAUSMARCHEN.    By  G.  E.  Pasnacht.     2s.  6d. 
*HAUFF— DIE  KARA  VANE.      By  Hebman  Hageb,  Ph.D.,   Lecturer  in  the 

Owens  College,  Manchester.     3s. 

*SCHMID,  CHR.  von— H.  VON  EICHENFELS.    By  G.  E.  Pasnacht.     2s.  6d. 

MACMILLAN'S  FOREIGN  SCHOOL  CLASSICS. —Edited  by  G.  E.  Pasnacht.  ISrao. 

FREYTAG  (G.)-DOKTOR  LUTHER.    By  P.  Stobb,  M.A.,  Headmaster  of  the 

Modern  Side^  Merchant  Taylors'  School.  [In  preparation. 

*GOETHE— GOTZ  VON  BERLICHINGEN.    By  H.  A.  Bull,  M.A.,  Assistant 

Master  at  Wellington.     2s. 
*GOETHE— FAUST.    Pabt  I.,  followed  by  an  Appendix  on  Pabt  II.    By  Jane 
Lee,  Lecturer  in  German  Literature  at  Newnham  College,  Cambridge.    4s.  6d. 

*HEINE— SELECTIONS  PROM  THE  REISEBILDER  AND  OTHER  PROSE 
WORKS.     By  C.  Colbeck,  M.A.,  Assistant  Master  at  Harrow.     2s.  6d. 


GERMAN — MODERN  GREEK — ITALIAN — SPANISH    21 

LESSING— MINNA  VON  BARNHELM.     By  James  Sime,  M. A.  [In  preparation. 
*SCHILLER— SELECTIONS  FROM  SCHILLER'S  LYRICAL  POEMS.    With  a 
Memoir  of  Schiller.     By  E.  J.  Turner,  B.A.,  and  B.  D.  A.  Morshead,  M.A., 
Assistant  Masters  at  Winchester.     2s.  6d. 
^SCHILLER— DIB  JUNGPRAU  VON  ORLEANS.  By  Joseph  Gostwick.  2s.  6d. 
*SCHILLBR  —MARIA  ST  U ART.    By  0.  Sheldon,  D.  Litt, ,  of  the  Royal  Academ- 
ical Institution,  Belfast.     2s.  6d. 
*SCHILLER— WILHELM  TELL.    By  G.*  E.  Fasnacht.     2s.  6d. 
*SCHILLBR— WALLBNSTEIN.     Part  I.  DAS  LAGER.    By  H.  B.  Cotterill, 

M.A.     2s. 
*UHLAND— SELECT  BALLADS.    Adapted  as  a  First  Easy  Reading  Book  for 

Beginners.    With  Vocabulary.     By  G.  E.  Fasnacht.     Is. 
*PYLODET.— NEW  GUIDE  TO  GERMAN  CONVERSATION ;  containing  an  Alpha- 
betical List  of  nearly  800  Familiar  Words ;  folloAved  by  Exercises,  Vocabulary 
of  Words  in  frequent  use.  Familiar  Phrases  and  Dialogues,  a  Sketch  of  German 
Literature,  Idiomatic  Expressions,  etc.     By  L.  Pylodet.    ISmo.    28.  6d. 
WHITNEY.— A  COMPENDIOUS  GERMAN  GRAMMAR.     By  W.  D.  Whitney, 
Professor  of  Sanskrit  and  Instructor  in  Modem  Languages  in  Yale  College. 
Cr.  8vo.     4s.  6d. 
A  GERMAN  READER  IN  PROSE  AND  VERSE.     By  the  same.     With  Notes 
and  Vocabulary.     Cr.  Svo.     5s. 
^WHITNEY   and   EDGREN.— A  COMPENDIOUS  GERMAN  AND  ENGLISH 
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Prof.  W.  D.  Whitney,  assisted  by  A.  H.  Edgren.    Cr.  Svo.    7s.  6d. 
THE  GBRMAN-ENGLISH  PART,  separately,  5s. 

MODERN  GREEK. 

VINCENT  and  DICKSON.— HANDBOOK  TO  MODERN  GREEK.  By  Sir  Edoar 
Vincent,  K.C.M.G.,  and  T.  G.  Dickson,  M.A.  With  Appendix  on  the  relation 
of  Modern  and  Classical  Greek  by  Prof.  Jebb.    Cr.  Svo.    6s. 


ITALIAN. 

DANTE.— THE  PURGATORY  OF  DANTE.  With  Translation  and  Notes,  by  A.  J. 
Butler,  M.A.    Cr:  Svo.    12s.  6d. 
THE.  PARADISO  OP  DANTE.     With  Translation  and  Notes,  by  the  same. 

2d.  Ed.     Cr.  Svo.     12s.  6d. 
READINGS  ON  THE  PURGATORIO  OP  DANTE.     Chiefly  based  on  the  Com- 
mentary of  Benvenuto  Da  Imola.    By  the  Hon.  W.  Warren  Vernon,  M.A. 
With  an  Introduction  by  the  Very  Rev.  the  Dean  of  St.  Paul's.    2  vols. 
Cr.  Svo.    24s. 

SPANISH. 

OALDERON.— POUR  PLAYS  OP  CALDERON.    With  Introduction  and  Notes. 
By  Norman  MacColl,  M.A.    Cr.  Svo.    14s. 
The  four  plays  here  given  are  El  Principe  Constante,  La  Vicla  es  Sueno,  El  Alcaldr. 
de  Zalamea,  and  El  Escondido  y  La  Tapada, 


22  MATHEMATICS 


MATHEMATICS. 

Arithmetic,  Book-keeping,  Algebra,  Euclid  and  Pure  Geometry,  Geometrical 
Drawing,  Mensuration,  Trigonometry,  Analytical  Geometry  (Plane  and 
Solid),  Problems  and  Question*  in  Mathematics,  Higher  Pure  Mathe- 
matics, Mechanics  (Statics,  Dynamics,  Hydrostatics,  Hydrodynamics :  see 
also  Physics),  Physics  (Sound,  Light,  Heat,  Electricity,  Elasticity,  Attrac- 
tions, &c.),  Astronomy,  Historical. 

ARITHMETIC. 

*ALDIS.— THE  GREAT  GIANT  ARITHMOS.  A  most  Elementary  Arithmetic 
for  Children.     By  Mary  Steadman  Alms.     Illustrated.     Gl.  8vo.     2s.  6d. 

ARMY  PRELIMINARY  EXAMINATION,  SPECIMENS  OF  PAPERS  SET  AT 
THE,  1882-89.— With  Answers  to  the  Mathematical  Questions.  Subjects: 
Arithmetic,  Algebra,  Euclid,  Geometrical  Drawing,  Geography,  French, 
English  Dictation.     Cr.  8vo.     3s.  6d. 

*BRADSHAW.— A  COURSE  OF    EASY  ARITHMETICAL   EXAMPLES   FOR 

BEGINNERS.     By  J.  G.  Bradshaw,  B.A.,  Assistant  Master  at  Clifton  College. 

Gl.  8vo.     2s.     With  Answers,  2s.  6d. 
*BROOKSMITH.— ARITHMETIC  IN  THEORY  AND  PRACTICE.    By  J.  Brook- 

SMITH,  M.A.    Cr.  Svo.    4s.  6d.     KEY.     Crown  Svo,    10s.  6d. 
*BROOKSMITH.— ARITHMETIC  FOR  BEGINNERS.    By  J.  and  E.  J.  Bbook- 

SMITH.     Gl.  Svo.     Is.  6d. 
CANDLER.— HELP  TO  ARITHMETIC.     Designed  for  the  use  of  Schools.    By  H. 

Candler,  Mathematical  Master  of  Uppingham  School.    2d  Ed.    Ex.  fcap.  Svo. 

2s.  6d. 
*DALTON.— RULES  AND  EXAMPLES  IN  ARITHMETIC.    By  the  Rev.  T.  Dal- 

TON,  M.A.,  Assistant  Master  at  Eton.    New  Ed.,  with  Answers.    ISmo.    2s.  6d. 
*GOYEN.— HIGHER  ARITHMETIC   AND   ELEMENTARY  MENSURATION. 

By  P.  GoYEN,  Inspector  of  Schools,  Dunedin,  New  Zealand.     Cr.  Svo.     5s. 

*HALL  and  KNIGHT.— ARITHMETICAL  EXERCISES  AND  EXAMINATION 
PAPERS.  With  an  Appendix  containing  Questions  in  Logarithms  and 
Mensuration.  By  H.  S.  Hall,  M.A.,  Master  of  the  Military  and  Engineering 
Side,  Clifton  College,  and  S.  R.  Knight,  B.A.     Gl.  Svo.     2s.  6d. 

LOCK.— Works  by  Rev.  J.  B.  Lock,  M.A.,  Senior  Fellow  and  Bursar  of  Gonville 
and  Caius  College,  Cambridge. 
^ARITHMETIC  FOR  SCHOOLS.     With  Answers  and  1000  additional  Examples 

for  Exercise.    3d  Ed.,  revised.     Gl.  Svo.     4s.  6d.     KEY.     Cr.  Svo.     10s.  6d. 
*ARITHMETIC  FOR  BEGINNERS.    A  School  Class-Book  of  Commercial  Arith- 
metic.    Gl.  Svo.     2s.  6d.     KEY.     Cr,  Svo.     8s.  6d. 
*A  SHILLING  BOOK  OF  ARITHMETIC,  FOR  ELEMENTARY  SCHOOLS. 
ISmo.    Is.     With  Answers.    Is.  6d. 
*PEDLEY.— EXERCISES  IN  ARITHMETIC  for  the  Use  of  Schools.    Containing 
more  than  7000  original  Examples.     By  Samuel  Pedley.    Cr.  8vo.    5s. 
Also  in  Two  Parts,  2s.  6d.  each. 
SMITH.— Works  by  Rev.  Barnard  Smith,  M.A.,  late  Fellow  and  Senior  Bursar  of 
St.  Peter's  College,  Cambridge. 
ARITHMETIC    AND   ALGEBRA,  in  their  Principles  and  Application ;  with 
numerous  systematically  arranged  Examples  taken  from  the  Cambridge  Exam- 
ination Papers,  with  especial  reference  to  the  Ordinary  Examination  for  the 
B.A.  Degree.    New  Ed.,  carefully  revised.    Cr.  Svo.    10s.  6d. 
»ARITHMETIC  FOR  SCHOOLS.     Cr.  Svo.    43.  6d.      KEY.    Cr,  Svo.    Ss.  6d. 


BOOK-KEEPING — ALGEBRA  23 

EXERCISES  IN  ARITHMETIC.     Cr.  8vo.     2s.     With  Answers,  2s.  6d.     An- 
swers separately,  6d. 

SCHOOL  GLASS-BOOK  OF  ARITHMETIC.     18mo.     3s.      Or  separately,  in 
Three  Parts,  Is.  each.     KEYS.     Parts  L,  IL,  and  IIL,  2s.  6d.  each. 

SHILLING  BOOK  OP  ARITHMETIC.     18mo.     Or  separately.  Part  I.,  2d. ; 

Part  II.,  3d. ;  Part  III.,  7d.    Answers,  6d.     KEY.     18mo.    4s.  6d. 
*THE  SAME,  with  Answers.    18mo,  cloth.     Is.  6d. 

EXAMINATION  PAPERS  IN  ARITHMETIC.     ISmo.     Is.  6d.     The  Same, 
with  Answers.     ISmo.     2s.    Answers,  6d.     KEY.     ISrao.     4s.  6d. 

THE  METRIC  SYSTEM  OF  ARITHMETIC,  ITS  PRINCIPLES  AND  APPLI- 
CATIONS, with  Numerous  Examples.     ISmo.     3d. 

A  CHART  OF  THE  METRIC  SYSTEM,  on  a  Sheet,  size  42  in.  by34  in.  on 
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EASY  LESSONS  IN  ARITHMETIC,  combining  Exercises  in  Reading,  Writing, 
Spelling,  and  Dictation.     Part  I.    Cr.  Svo.    9d. 

EXAMINATION  CARDS  IN  ARITHMETIC.     With  Answers  and  Hints. 

Standards  I.  and  II.,  in  box.  Is.    Standards  III.,  IV.,  and  V.,  in  boxes.  Is.  each. 
Standard  VI.  in  Two  Parts,  in  boxes,  Is.  each, 

A  and  B  papers,  of  nearly  the  same  difficulty,  are  given  so  as  to  prevent  copying, 
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BOOK-KEEPING-. 

*THORNTON.— FIRST  LESSONS  IN  BOOK-KEEPING.    By  J.  Thornton.     Cr. 
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*PRIMER  OF  BOOK-KEEPING.     18mo.     Is.     KEY.     Demy  Svo.     2s.  6d. 

ALG-EBRA. 

*DALTON.— RULES  AND  EXAMPLES  IN  ALGEBRA.     By  Rev.  T.  Dalton, 
Assistant  Master  at  Eton.     Part  I.     ISmo.    2s.      KEY.     Cr.  Svo.     78.  6d. 
Part  II.     ISmo,     2s.  6d. 
HALL  and  KNIGHT.— Works  by  H.  S.  Hall,  M.A.,  Master  of  the  Military  and 

Engineering  Side,  Clifton  College,  and  S.  R.  Knight,  B.A. 
*ELEMENTARY  ALGEBRA  FOR  SCHOOLS.    6th  Ed.,  revised  and  corrected. 
Gl.  Svo,  bound  in  maroon  coloured  cloth,  3s.  6d. ;  with  Answers,  bound  in 
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^ALGEBRAICAL  EXERCISES  AND  EXAMINATION  PAPERS.     To  accom- 
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*HIGHER  ALGEBRA.     3d  Ed.    Cr.  Svo.     7s.  6d.     KEY.    Cr.  Svo.    10s.  6d. 
*JONES    and    CHEYNE.— ALGEBRAICAL   EXERCISES.       Progressively  Ar- 
ranged.   By  Rev.  C.  A.  Jones  and  0.  H.  Cheyne,  M.A.,  late  Mathematical 
Masters  at  Westminster  School.    ISmo.    2s.  6d. 
KEY.    By  Rev.  W.  Failes,  M.A.,  Mathematical  Master  at  Westminster  School. 
Cr.  Svo.     7s.  6d. 
SMITH.— ARITHMETIC  AND  ALGEBRA,  in  their  Principles  and  Application ; 
with  numerous  systematically  arranged  Examples  taken  from  the  Cambridge 
Examination  Papers,  with  especial  reference  to  the  Ordinary  Examination  for 
the  B.A.  Degree.     By  Rev.  Barnard  Smith,  M.A.     New  Edition,  carefully 
revised.     Cr.  Svo.     10s.  6d. 
SMITH.— Works  by  Charles    Smith,  M.A.,    Master  of  Sidney  Sussex  College, 

Cambridge. 
*ELEMENTARY  ALGEBRA.    2d  Ed.,  revised.    Gl.  Svo.    43.  6d. 
*A  TREATISE  ON  ALGEBRA.   2d  Ed.  Cr.  Svo.   7s.  6d.     KEY,   Cr.Svo.   10.s,  6d. 
TODHUNTER.— Works  by  Isaac  Todhunter,  F.R.S. 
*ALGBBRA  FOR  BEGINNERS.     ISmo.    28.  6d.     KEY.    Cr.  Svo.     6s.  6d. 


24  MATHEMATICS 

*ALQEBRA  FOR  COLLEGES  AND  SCHOOLS.  Cr.  8vo.  7s.  6d.  KEY.  Cr. 

8vo,  10s.  6d. 

EUCLID  AND  PURE  GEOMETRY. 

COOKSHOTT  and  WALTERS.— A  TREATISE  ON  GEOMETRICAL  CONICS. 
In  accordance  with  the  Syllabus  of  the  Association  for  the  Improvement  of 
Geometrical  Teaching.  By  A.  Cockshott,  M.A.,  Assistant  Master  at  Eton, 
and  Rev.  F-.  B.  Walters,  M.A.,  Principal  of  King  William's  College,  Isle  of 
Man.     Cr.  8vo.     6s. 

CONSTABLE.— GEOMETRICAL  EXERCISES  FOR  BEGINNERS.  By  Samuel 
Constable.     Cr.  Svo.     3s.  6d. 

OUTHBERTSON.— EUCLIDIAN  GEOMETRY.  By  Francis  Cuthbertson,  M.A., 
LL.D.     Ex.  fcap.  Svo.    4s.  6d. 

DAY.— PROPERTIES  OF  CONIC  SECTIONS  PROVED  GEOMETRICALLY. 
By  Rev.  H.  G.  Day,  M.A.  Part  I.  The  Ellipse,  with  an  ample  collection  of 
Problems.     Cr.  Svo.     8s.  6d. 

DEAKIN.— RIDER  PAPERS  ON  EUCLID,  BOOKS  I.  and  IL  By  Rupert 
Deakin,  M.A.     ISmo.     Is. 

DODGSON.— Works  by  Charles  L.  Dodgson,  M.A.,  Student  and  late  Mathematical 
Lecturer,  Christ  Church,  Oxford. 
EUCLID,  BOOKS  I.  and  II.     6th  Ed.,  with  words  substituted  for  the  Alge- 
braical Symbols  used  in  the  1st  Ed.     Cr.  Svo.     2s. 
EUCLID  AND  HIS  MODERN  RIVALS.    2d  Ed.    Cr,  Svo.    6s. 
CURIOSA  MATHEMATICA.     Part  I.     A  New  Theory  of  Parallels.     3d  Ed. 
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DREW.— GEOMETRICAL  TREATISE  ON  CONIC  SECTIONS.  By  W.  H. 
Drew,  M.A.    New  Ed.,  enlarged.     Cr.  Svo.    5s. 

DUPUIS.— ELEMENTARY  SYNTHETIC  GEOMETRY  OF  THE  POINT,  LINE 
AND  CIRCLE  IN  THE  PLANE.  By  N.  F.  Dupuis,  M.A.,  Professor  of  Pure 
Mathematics  in  the  University  of  Queen's  College,  Kingston,  Canada.  Gl.  Svo. 
4s.  6d. 

*HALL  and  STEVENS,— A  TEXT-BOOK  OF  EUCLID'S  ELEMENTS.  In- 
eluding  Alternative  Proofs,  together  with  additional  Theorems  and  Exercises, 
classified  and  arranged.  By  H.  S.  Hall,  M.A.,  and  F.  H.  Stevens,  M.A., 
Masters  of  the  Military  and  Engineering  Side,  Clifton  College.  Gl.  Svo.  Book 
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HALSTED.— THE  ELEMENTS  OF  GEOMETRY.  By  G.  B.  Halsted,  Professor 
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HAYWARD.— THE  ELEMENTS  OP  SOLID  GEOMETRY,  By  R.  B,  Hayward, 
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LOOK,— EUCLID  FOR  BEGINNERS.  Being  an  Introduction  to  existing  Text- 
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MILNE  and  DAVIS,— GEOMETRICAL  CONICS,  Part  I,  The  Parabola,  By 
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RICHARDSON.— THE  PROGRESSIVE  EUCLID.  Books  I.  and  IL  With  Notes, 
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SYLLABUS  OF  PLANE  GEOMETRY  (corresponding  to  Euclid,  Books  I.-VI,)— 
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*TODHUNTER.— THE  ELEMENTS  OF  EUCLID.  By  I.  Todhunter,  F.R.S. 
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WILSON.— Works  by  Ven.  Archdeacon  Wilson,  M.A.,  formerly  Headmaster  of 
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GEOMETRICAL  DRAWING — TRIGONOMETRY  25 

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TRIGONOMETRY. 

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JOHNSON.— A  TREATISE  ON  TRIGONOMETRY.  By  W.  E.  Johnson,  M.A., 
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26  MATHEMATICS 

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MATTHEWS.— MANUAL  OF  LOGARITHMS.  By  G.  F.  Matthews,  B.A.  8vo. 
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port.     Gl.  Svo.     4s.  6d. 

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PROBLEMS  AND  QUESTIONS  IN 
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28  MATHEMATICS 

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MECHANICS  29 

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JELLETT.— A  TREATISE  ON  THE  THEORY  OF  FRICTION.     By  John  H. 

Jellktt,  B.D.,  late  Provost  of  Trinity  College,  Dublin.    Svo.    8s.  6d. 
KENNEDY.— THE  MECHANICS  OF  MACHINERY.     By  A.  B.  W.  Kennedy, 

F.R.S.     Illustrated.     Cr.  Svo.     12s.  6d. 
LOOK.— Works  by  Rev.  J.  B.  Lock,  M.A. 
*BLEMENTARY  STATICS.     2d  Ed.     Gl.  Svo.     4s.  6d. 
♦ELEMENTARY  DYNAMICS.     3d  Ed.    Gl.  Svo.    4s.  6d. 
MECHANICS  FOR  BEGINNERS.     Gl.  Svo.  [In  the  Press. 

MACGREGOR.— KINEMATICS  AND  DYNAMICS.  An  Elementary  Treatise. 
By  J.  G.  MacGregor,  D.Sc,  Munro  Professor  of  Physics  in  Dalhousie  College, 
Halifax,  Nova  Scotia.  Illustrated.  Cr.  Svo.  IDs.  6d, 
PARKINSON.— AN  ELEMENTARY  TREATISE  ON  MECHANICS.  By  S. 
Parkinson,  D.D.,  F.R.S.,  late  Tutor  and  Prselector  of  St.  John's  College, 
Cambridge.  6th  Ed.,  revised.  Cr.  Svo.  9s.  6d. 
PIRIE.— LESSONS  ON  RIGID  DYNAMICS.    By  Rev.  G.  Pirie,  M.A.,  Professor 

of  Mathematics  in  the  University  of  Aberdeen.     Cr.  Svo.     6s. 
ROUTE.— Works  by  Edward  John  Routh,  D.Sc,  LL.D.,  F.R.S.,  Hon.  Fellow 
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A  TREATISE  ON  THE  DYNAMICS  OF  THE  SYSTEM  OF  RIGID  BODIES. 
With  numerous  Examples.     Two  Vols.     Svo.     Vol.   I.— Elementary  Parts. 
5th  Ed.     14s.    Vol.  IL— The  Advanced  Parts.    4th  Ed.    14s. 
STABILITY  OF  A  GIVEN  STATE  OP  MOTION,  PARTICULARLY  STEADY 
MOTION.    Adams  Prize  Essay  for  1S77.    Svo.    Ss.  6d. 
♦SANDERSON.- HYDROSTATICS  FOR  BEGINNERS.     By  P.  W.  Sanderson, 

M.A.,  Assistant  Master  at  Dulwich  College.    Gl.  Svo.    4s.  6d. 
TAIT  and  STEELE.— A  TREATISE  ON  DYNAMICS  OF  A  PARTICLE.      By 
Professor  Tait,  M.A,  and  W.  J.  Steele,  B.A   6th  Ed.,  revised.   Cr.  Svo.   12s. 
TODHUNTER.— Works  by  Isaac  Todhunter,  F.R.S. 
^MECHANICS  FOR  BEGINNERS.     ISmo.    4s.  6d.     KEY.    Cr.  Svo.    6s.  6d. 
A  TREATISE  ON  ANALYTICAL  STATICS.    5th  Ed.     Edited  by  Prof.  J.  D. 
Everett,  F.R.S.     Cr.  Svo.    10s.  Gd. 


30  MATHEMATICS 

PHYSIOS :  Sound,  Light,  Heat,  Electricity,  Elasticity, 
Attractions,  etc.     (See  also  Mechanics.) 

AIRY.— Works  by  Sir  G.  B.  Aihy,  K.C.B.,  formerly  Astronomer-Royal. 
ON  SOUND  AND  ATMOSPHERIC  VIBRATIONS.      With  the  Mathematical 

Elements  of  Music.     2d  Ed.,  revised  and  enlarged.     Or.  8vo.     9s. 
GRAVITATION :  An  Elementary  Explanation  of  the  Principal  Perturbations  in 
the  Solar  System.     2d  Ed.     Or.  8vo.     7s.  6d. 

CLAUSIUS.— MECHANICAL  THEORY  OP  HEAT.     By  R.  Clausius.     Trans- 
lated by  W.  R.  Bkowne,  M.A.     Cr.  8vo.     10s.  6d. 
CmvnVTTNG.— AN    INTRODUCTION    TO   THE   THEORY   OF   ELECTRICITY. 

By  LiNNiEUS  CuMMiNO,  M.A. ,  Assistant  Master  at  Rugby.    Illustrated.   Or.  Svo. 

8s.  6d. 
DANIELL.— A  TEXT-BOOK  OF  THE  PRINCIPLES  OF  PHYSICS.    By  Alfred 

DAi^iELL,  D.Sc.     Illustrated.     2d  Ed.,  revised  and  enlarged.     8vo.     21s. 
DAY.— ELECTRIC  LIGHT  ARITHMETIC.    By  R.  E.  Day,  Evening  Lecturer  in 

Experimental  Physics  at  King's  College,  London.     Pott  Svo.     2s. 
EVERETT.— UNITS  AND  PHYSICAL  CONSTANTS.    By  J.  D.  Everett,  P.R.S., 

Professor  of  Natural  Philosophy,  Queen's  College,  Belfast.    2d  Ed.     Ex.  fcap. 

8vo.     5s. 
FERRERS.— AN  ELEMENTARY  TREATISE  ON  SPHERICAL  HARMONICS, 

and  Subjects  connected  with  them.     By  Rev.  N.  M.  Ferrers,  D.D.,  F.R.S., 

Master  of  Gonville  and  Caius  College,  Cambridge.     Cr.  8vo.    7s.  6d. 
FESSENDEN.  — PHYSICS    FOR    PUBLIC     SCHOOLS.      By    C.    Fessenden. 

Illustrated.    Fcap.  8vo.  [In  the  Press. 

GRAY.— THE  THEORY  AND  PRACTICE  OF  ABSOLUTE  MEASUREMENTS 

IN  ELECTRICITY  AND  MAGNETISM.     By  A.  Gray,  F.R.S.E.,  Professor 

of  Physics  in  the  University  College  of  North  Wales.    Two  Vols.     Cr.  Svo. 

Vol.  I.     12s.  6d.  [Vol.  IL    In  the  Press. 

ABSOLUTE  MEASUREMENTS  IN  ELECTRICITY  AND  MAGNETISM.     2d 

Ed.,  revised  and  greatly  enlarged.     Fcap.  Svo.     5s.  6d. 
IBBETSON.— THE  MATHEMATICAL   THEORY  OP  PERFECTLY  ELASTIC 

SOLIDS,  with  a  Short  Account  of  Viscous  Fluids.     By  W.  J.  Ibbetson,  late 

Senior  Scholar  of  Clare  College,  Cambridge.    Svo.     21s. 
*JONES.— EXAMPLES  IN  PHYSICS.      Containing  over  1000  Problems  with 

Answers  and  numerous  solved  Examples.    Suitable  for  candidates  preparing 

for  the  Intermediate,  Science,  Preliminary,  Scientific,  and  other  Examinations 

of  the  University  of  London.    By  D.  E.  Jones,  B.Sc,  Professor  of  Physics 

in  the  University  College  of  Wales,  Aberystwyth.     Fcap.  Svo.     3s.  6d. 
*ELEMENTARY  LESSONS  IN  HEAT,  LIGHT,  AND  SOUND.     Gl.  Svo.    2s.  6d 
LOOKYER.— CONTRIBUTIONS  TO  SOLAR  PHYSICS.     By  J.  Norman  Lockyer 

F.R.S.     With  Illustrations.     Royal  Svo.     31s.  6d. 
LODGE.— MODERN  VIEWS  OF  ELECTRICITY.    By  Oliver  J.  Lodge,  F.R.S. . 

Professor  of  Experimental  Physics  in  University  College,  Liverpool.     lUus' 

trated.     Cr.  Svo.     6s.  6d. 
LOEWY.— *QUESTIONS  AND  EXAMPLES  ON  EXPERIMENTAL  PHYSICS 

Sound,  Light,  Heat,  Electricity,  and  Magnetism.     By  B.  Loewy,  Examiner  in 

Experimental  Physics  to  the  College  of  Preceptors.     Fcap.  Svo.     2s. 
*A  GRADUATED  COURSE  OF  NATURAL  SCIENCE  FOR  ELEMENTARY 

AND  TECHNICAL  SCHOOLS  AND  COLLEGES.     By  the  same.     In  Three 

Parts.    Part  I.  First  Year's  Course.     Gl.  Svo.    2s. 
LUPTON.— NUMERICAL    TABLES    AND    CONSTANTS    IN    ELEMENTARY 

SCIENCE.     By  S.  Lupton,  M.A.,  late  Assistant  Master  at  Harrow.    Ex.  fcap. 

Svo.     2s.  6d. 
MACFARLANE.— PHYSICAL  ARITHMETIC.     By  A.  Macfarlane,  D.Sc,  late 

Examiner  in  Mathematics  at  the  University  of  Edinburgh.    Cr.  Svo.     7s.  6d. 


PHYSICS  81 

*MAYER.— SOUND :  A  Series  of  Simple,  Entertaining,  and  Inexpensive  Experi- 
ments in  the  Phenomena  of  Sound.  By  A.  M.  Mayer,  Professor  of  Physics 
in  the  Stevens  Institute  of  Technology.     Illustrated.     Cr.  8vo.    3s.  6d. 

*MAYER  and  BARNARD.— LIGHT :  A  Series  of  Simple,  Entertaining,  and  In- 
expensive Experiments  in  the  Phenomena  of  Light.  By  A.  M.  Mayer  and 
0.  Barnard.     Illustrated.     Cr,  8vo,     2s.  6d. 

MOLLOY.— GLEANINGS  IN  SCIENCE  :  Popular  Lectures  on  Scientific  Subjects. 
By  the  Rev.  Gerald  Molloy,  D.Sc,  Rector  of  the  Catholic  University  of 
Ireland.     8vo.     7s.  6d. 

NEWTON.— PRINCIPI A.     Edited  by  Prof.  Sir  W.  Thomson,  P.R.S.,  and  Prof. 

Blackburnb.    4to.     31s.  6d. 

THE  FIRST  THREE  SECTIONS  OF  NEWTON'S  PRINCIPIA.     With  Notes 

and  Illustrations.     Also  a  Collection  of  Problems,  principally  intended  as 

Examples  of  Newton's  Methods.    By  P.  Frost,  M.A.,  D.Sc.    3d  Ed.    Svo.    12s. 

PARKINSON.— A  TREATISE  ON  OPTICS.  By  S.  Parkinson,  D.D.,  F.R.S., 
late  Tutor  and  Prselector  of  St.  John's  College,  Cambridge.  4th  Ed.,  revised 
and  enlarged.     Cr.  Svo.    10s.  6d. 

PEABODY.- THERMODYNAMICS  OP  THE  STEAM-ENGINE  AND  OTHER 
HEAT-ENGINES.  By  Cecil  H.  Peabody,  Associate  Professor  of  Steam 
Engineering,  Massachusetts  Institute  of  Technology,     Svo.     21s. 

PERRY,  —  STEAM :  An  Elementary  Treatise.  By  John  Perry,  Professor 
of  Mechanical  Engineering  and  Applied  Mechanics  at  the  Technical  College, 
Finsbury.     ISmo.     4s.  6d, 

PICKERING.— ELEMENTS  OF  PHYSICAL  MANIPULATION.  By  Prof.  Ed- 
ward C.  Pickering.    Medium  Svo.     Part  I.,  12s.  6d.    Part  II.,  14s. 

PRESTON.— THE  THIiPRY  OF  LIGHT,     By  Thomas  Prkston,  M.A,     Illus- 
trated.    Svo.    12s,  6d. 
THE  THEORY  OF  HEAT,    By  the  same  Author.    Svo,  [In  preparation. 

RAYLEIGH,— THE  THEORY  OF  SOUND,  By  Lord  Rayleigh,  F.R.S.  Svo. 
Vol.  I.,  12s.  6d.    Vol.  II.,  12s.  6d.  [Vol.  III.     In  the  Press. 

SHANN.— AN  ELEMENTARY  TREATISE  ON  HEAT,  IN  RELATION  TO 
STEAM  AND  THE  STEAM-ENGINE,  By  G,  Shann,  M.A,  Illustrated. 
Cr.  Svo.     4s.  6d, 

SPOTTISWOODE.— POLARISATION  OF  LIGHT,  By  the  late  W.  Spottiswoode, 
F.R.S.     Illustrated.     Cr,  Svo.     3s.  6d, 

STEWART,— Works  by  Balfour  Stewart,  F.R.S.,  late  Langworthy  Professor  of 

Physics  in  the  Owens  College,  Victoria  University,  Manchester. 
*PRIMER  OF  PHYSICS.     Illustrated.     With  Questions.     ISmo.     Is, 
*LESSONS  IN  ELEMENTARY  PHYSICS.     Illustrated.     Fcap,  Svo,    4s,  6d, 
*QUESTIONS.     By  Prof.  T,  H,  Core,     Fcap.  Svo.    2s. 

STEWART  and  GEE,— LESSONS  IN  ELEMENTARY  PRACTICAL  PHYSICS. 
By  Balfour  Stewart,  F.R.S.,  and  W,  W,  Haldane  Gee,  B.Sc.  Cr.  Svo. 
Vol,  I,  General  Physical  Processes,  6s.  Vol,  II.  Electricity  and 
Magnetism.    7s.  6d,  [Vol.  III.  Optics,  Heat,  and  Sound.    In  the  Press. 

^PRACTICAL  PHYSICS  FOR  SCHOOLS  AND  THE  JUNIOR  STUDENTS  OF 
COLLEGES.     Gl,  Svo.     Vol.  I,  Electricity  and  Magnetism.     2s.  6d. 

[Vol.  II,  Optics,  Heat,  and  Sound.    In  the  Press. 

STOKES.— ON  LIGHT.  Burnett  Lectures,  delivered  in  Aberdeen  in  1883-4-5. 
By  Sir  G.  G.  Stokes,  F.R.S.,  Lucasian  Professor  of  Mathematics  in  the 
University  of  Cambridge.  First  Course  :  On  the  Nature  of  Light,  Second 
Course :  On  Light  as  a  Means  of  Investigation,  Third  Course :  On  the 
Beneficial  Effects  of  Light,  Cr.  Svo,  7s.  6d. 
\*  The  2d  and  3d  Courses  may  be  had  separately,    Cr,  Svo,    2s,  6d,  each, 

STONE,— AN  ELEMENTARY  TREATISE  ON  SOUND.  By  W.  H.  Stone, 
Illustrated,    Fcap.  Svo.    3s.  6d. 

TAIT.— HEAT,    By  P.  G,  Tait,  Professor  of  Natural  Philosophy  in  the  University 
of  Edinburgh.     Cr.  Svo.     6s. 
LECTURES  ON  SOME  RECENT  ADVANCES  IN  PHYSICAL  SCIENCE.     By 
the  same.    3d  Edition,     Crown  Svo,    9s. 


82  MATHEMATICS 

TAYLOR.— SOUND  AND  MUSIC.    An  Blemeutaiy  Treatise  on  the  Physical  Con- 

stitution  of  Musical  Sounds  and  Harmony,  including  the  Chief  Acoustical 

Discoveries  of  Professor  Helmholtz,     By  Sedley  Taylob,  M.A.     Illustrated. 

2d  Ed.     Ex.  cr.  8vo.    8s.  6d. 

*THOMPSON.  — ELEMENTARY  LESSONS  IN  ELECTRICITY  AND  MAGNET- 

ISM.     By  SiLVANUS  P.  Thompson,  Principal  and  Professor  of  Physics  in  the 

Technical  College,  Finsbury.  Illustrated.  New  Ed.,  revised.   Fcap.  8vo.  4s.  6d. 

THOMSON.— Works  by  J.  J.  Thomson,  Professor  of  Experimental  Physics  in  the 

University  of  Cambridge. 

A  TREATISE  ON  THE  MOTION  OF  VORTEX  RINGS.    Adams  Prize  Essay, 

1882.     8vo.    6s. 
APPLICATIONS  OF  DYNAMICS  TO  PHYSICS  AND  CHEMISTRY.     Cr.  8vo. 
7s.  6d. 
THOMSON.— Works  by  Sir  W.  Thomson,  P.R.S.,  Professor  of  Natural  Philosophy 
in  the  University  of  Glasgow. 
ELECTROSTATICS    AND    MAGNETISM,    REPRINTS    OF    PAPERS    ON. 

2d  Ed.    8vo.    18s. 
POPULAR  LECTURES  AND  ADDRESSES.     3  Vols.     lUustrated.     Cr.  Svo. 
Vol.  I.  Constitution  of  Matter.    6s.    Vol.  III.  Navigation.      [In  the  Press. 
TODHUNTER.— Works  by  Isaac  Todhunter,  F.R.S. 
AN  ELEMENTARY  TREATISE  ON  LAPLACE'S,  LAME'S,  AND  BESSBL'S 
-    FUNCTIONS.     Crown  Svo.     10s.  6d. 

A  HISTORY  OF  THE  MATHEMATICAL  THEORIES  OF  ATTRACTION,  AND 
THE  FIGURE  OF  THE  EARTH,  from  the  time  of  Newton  to  that  of  Laplace. 
2  vols.    Svo.     24s. 
TURNER.— A  COLLECTION  OF  EXAMPLES  ON  HEAT  AND  ELECTRICITY. 

By  H.  H.  Turner,  Fellow  of  Trinity  College,  Cambridge.     Cr.  Svo.     2s.  6d. 
WRIGHT.— LIGHT :  A  Course  of  Experimental  Optics,  chiefly  with  the  Lantern. 
By  Lewis  Wright.    Illustrated.    Cr.  8vo.    7s.  6d. 

ASTRONOMY. 

AIRY.— Works  by  Sir  G.  B.  Airy,  K.C.B.,  formerly  Astronomer-Royal. 
*POPULAR  ASTRONOMY.     18mo.     4s.  6d. 

GRAVITATION :  An  Elementary  Explanation  of  the  Principal  Perturbations  in 

the  Solar  System.     2d  Ed.     Cr.  8vo.    7s.  6d. 

OHEYNE.— AN  ELEMENTARY  TREATISE  ON  THE  PLANETARY  THEORY. 

By  C.  H.  H.  Cheyne.    With  Problems.     3d  Ed.    Edited  by  Rev.  A.  Freeman, 

M.A,  F.R.AS.     Cr.  Svo.    7s.  6d. 

CLARK  and  SADLER.^THE  STAR  GUIDE.    By  L.  Clark  and  H.  Sadleb. 

Roy.  Svo.     5s. 
CROSSLEY,  GLEDHILL,  and  WILSON.— A  HANDBOOK  OF  DOUBLE  STARS. 
By  E.  Crossley,  J.  Gledhill,  and  J.  M.  Wilson.    Svo.    21s. 
CORRECTIONS  TO  THE  HANDBOOK  OF  DOUBLE  STARS.     Svo.    Is. 
FORBES.- TRANSIT  OF  VENUS.     By  G.  Forbes,  Professor  of  Natural  Philo- 
sophy in  the  Andersonian  University,  Glasgow.    Illustrated.    Cr.  Svo.    3s.  6d. 
(K)DFRAY. — Works  by  Hugh  Godfray,  M.  A.,  Mathematical  Lecturer  at  Pembroke 
College,  Cambridge. 
A  TREATISE  ON  ASTRONOMY.     4th  Ed.     Svo.    12s.  6d. 
AN  ELEMENTARY  TREATISE  ON  THE  LUNAR  THEORY,  with  a  brief 
Sketch  of  the  Problem  up  to  the  time  of  Newton.    2d  Ed.,  revised.    Cr.  Svo. 
5s.  6d. 
LOOKYER.— Works  by  J.  Norman  Lockyer,  F.R.S. 
*PRIMER  OF  ASTRONOMY.    Illustrated.     ISrao.     Is. 

*ELEMENTARY  LESSONS  IN  ASTRONOMY.    With  Spectra  of  the  Sun,  Stars, 
and  Nebulae,  and  numerous  Illustrations.    36th  Thousand.    Revised  through- 
out.    Fcap.  Svo.     5s.  6d. 
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By  J.  Forbes  Robertson.    ISmo,     Is.  6d. 


ASTRONOMY — HISTORICAL  33 

THE  CHEMISTRY  OF  THE  SUN.    lUustrated.     8vo.    14s. 

THE    METEORITIO    HYPOTHESIS    OP    THE    ORIGIN    OF    COSMICAL 
SYSTEMS.     Illustrated.    8vo.     17s.  net. 

THE  EVOLUTION  OF  THE  HEAVENS  AND  THE  EARTH.  Cr.  8vo.  Illus- 
trated. [In  the  Press. 
LOCKYER  and  SEABROKE.— STAR-GAZING  PAST  AND  PRESENT.  By  J. 
Norman  Lockyer,  F.R.S.  Expanded  from  Shorthand  Notes  with  the 
assistance  of  G.  M.  Seabroke,  F.R.A.S.  Royal  Svo.  21s. 
NEWCOMB.— POPULAR  ASTRONOMY.  By  S.  Newcomb,  LL.D.,  Professor 
U.S.  Naval  Observatory.     Illustrated.     2d  Ed.,  revised.    Svo.    18s. 

HISTOBICAL. 

BALL.— A  SHORT  ACCOUNT  OF  THE  HISTORY  OF  MATHEMATICS.     By  W. 

W.  R.  Ball,  M.A.     Cr.  8vo.     10s.  6d. 
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A  HISTORY  OF  THE  MATHEMATICAL  THEORY  OP  PROBABILITY,  from 

the  time  of  Pascal  to  that  of  Laplace.    8vo.    18s. 
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Laplace.     2  vols.     Svo.    24s. 


NATURAL   SCIENCES. 

Chemistry ;  Physical  Greography,  Geology,  and  Mineralogy ;  Biology ; 
Medicine. 

(For  MECHANICS,  PHYSICS,  and  ASTRONOMY,  see 
MATHEMATICS.) 

CHEMISTRY. 

ARMSTRONG.— A  MANUAL  OF  INORGANIC  CHEMISTRY.  By  Henry  Arm- 
strong, F.R.S.,  Professor  of  Chemistry  in  the  City  and  Guilds  of  London  Tech- 
nical Institute.     Cr.  Svo.  [In  preparation. 

»COHEN.— THE  OWENS  COLLEGE  COURSE  OF  PRACTICAL  ORGANIC 
CHEMISTRY.  By  Julius  B.  Cohen,  Ph.D.,  Assistant  Lecturer  on  Chemistry 
in  the  Owens  College,  Manchester.  With  a  Preface  by  Sir  Henry  Roscoe, 
F.R.S.,  and  C.  Schorlemmer,  F.R.S.     Fcap.  Svo.     2s.  Od. 

COOKE.— ELEMENTS  OF  CHEMICAL  PHYSICS.  By  Josiah  P.  Cooke,  Jun., 
Erving  Professor  of  Chemistry  and  Mineralogy  in  Harvard  University.  4th  Ed. 
Svo.     21s. 

FLEISCHER.— A  SYSTEM  OF  VOLUMETRIC  ANALYSIS.  By  Emil  Fleischer. 
Translated,  with  Notes  and  Additions,  by  M.  M.  P.  MuiR,  F.R.S.E.  Illustrated. 
Cr.  Svo.    Ts.  6d. 

FRANKLAND.— A  HANDBOOK  OF  AGRICULTURAL  CHEMICAL  ANALYSIS. 
By  P.  F.  Frank  LAND,  F.R.S.,  Professor  of  Chemistry  in  University  College, 
Dundee.     Cr.  Svo.     7s.  6d. 

HARTLEY.— A  COURSE  OF  QUANTITATIVE  ANALYSIS  FOR  STUDENTS. 
By  W.  Noel  Hartley,  F.R.S.,  Professor  of  Chemistry  and  of  Applied  Chemis- 
try, Science  and  Art  Department,  Royal  College  of  Science,  Dublin.  Gl. 
Svo.     5s. 

mORNS.— PRACTICAL  METALLURGY  AND  ASSAYING.  A  Text -Book  for 
the  use  of  Teachers,  Students,  and  Assayers.  By  Arthur  H.  Hiorns,  Prin- 
cipal of  the  School  of  Metallurgy,  Birmingham  and  Midland  Institute.  Illus- 
trated.   Gl.  Svo.    6s. 


84  NATURAL  SCIENCES 

A  TEXT-BOOK  OF  ELEMENTARY   METALLURGY  FOR  THE  USE    OF 
STUDENTS.    To  which  is  added  an  Appendix  of  Examination  Questions,  em- 
bracing the  whole  of  the  Questions  set  in  the  three  stages  of  the  subject  by  the 
Science  and  Art  Department  for  the  past  twenty  years.  By  the  same.  Gl.  8vo.  4s. 
IRON  AND  STEEL  MANUFACTURE.     A  Text-Book  for  Beginners.    By  the 

same.    Illustrated.     Gl.  8vo.     3s.  6d. 
MIXED  METALS  OR  METALLIC  ALLOYS.    By  the  same.    Gl.  8vo.     6s. 
JONES.— *THE  OWENS  COLLEGE  JUNIOR  COURSE  OF  PRACTICAL  CHEM- 
ISTRY.  By  Francis  Jones,  F.R.S.E.,  Chemical  Master  at  the  Grammar  School, 
Manchester.    With  Preface  by  Sir  Henry  Roscoe,  F.R.S.    Illustrated.    Fcap. 
Bvo.    2s.  6d. 
♦QUESTIONS  ON  CHEMISTRY.    A  Series  of  Problems  and  Exercises  in  Inorganic 
and  Organic  Chemistry.     By  the  same.     Fcap.  Svo.    3s.  ^ 
LAin)AUER.-:^BLOWPIPE  ANALYSIS.    By  J.  Landauer.    Authorised  English 
Edition  by  J.  Taylor  and  W.  E.  Kay,  of  Owens  College,  Manchester. 

[New  Edition  in  preparation. 
LOOKYER.— THE  CHEMISTRY  OP  THE  SUN.    By  J.  Norman  Lockyer,  F,R.S. 

Illustrated.    Svo.    14s. 
LUPTON.— CHEMICAL  ARITHMETIC.     With  1200  Problems.    By  S.  Lupton, 

M.A.     2d  Ed.,  revised  and  abridged.    Fcap.  Svo.     4s.  6d. 

MANSFIELD.— A  THEORY  OF  SALTS.     By  C.  B.  Mansfield.     Crown  Svo.   143. 

MELDOLA.— THE  CHEMISTRY  OF  PHOTOGRAPHY.    By  Raphael  Meldola, 

F.R.S.,  Professor  of  Chemistry  in  the  Technical  College,  Finsbury.  Cr.  Svo.  6s. 

MEYER.     HISTORY  OF  CHEMISTRY.     By  Ernst  von  Meyer.    Translated  by 

George  MoGowan,  Ph.D.     Svo.     14s.  net. 
MIXTER.— AN  ELEMENTARY  TEXT-BOOK  OF  CHEMISTRY.    By  William  G. 
MixTER,  Professor  of  Chemistry  in  the  ShefBeld  Scientific  School  of  Yale  College. 
2d  and  revised  Ed.    Cr.  Svo.     7s.  6d. 
MUIR.— PRACTICAL  CHEMISTRY  FOR  MEDICAL  STUDENTS.    Specially  ar- 
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lector  in  Chemistry  at  Gonville  and  Caius  College,  Cambridge.  Fcap.  Svo.  Is.  6d 
MUIR  and  WILSON.— THE  ELEMENTS  OF  THERMAL  CHEMISTRY.    By  M. 

M.  P.  Muir,  F.R.S.E.  ;  assisted  by  D.  M.  Wilson.    Svo.    12s.  6d. 
OSTWALD.— OUTLINES    OF    GENERAL    CHEMISTRY    (PHYSICAL   AND 
THEORETICAL).    By  Prof.  W.  Ostwald.     Translated  by  James  Walker, 
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RAMSAY.— EXPERIMENTAL  PROOFS  OF  CHEMICAL  THEORY  FOR  BE- 
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sity College,  London.    Pott  Svo.     2s.  6d. 
REMSEN.— Works  by  Ira  Remsen,  Professor  of  Chemistry  in  the  Johns  Hopkins 
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COMPOUNDS  OP  CARBON :  or,  Organic  Chemistry,  an  Introduction  to  the 

Study  of.    Cr.  Svo.    6s.  6d. 
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*THE  ELEMENTS  OF  CHEMISTRY.    A  Text-Book  for  Beginners.     Fcap.  Svo. 

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ROSCOE.— Works  by  Sir  Henry  E.  Roscoe,  F.R.S.,  formerly  Professor  of  Chemistry 

in  the  Owens  College,  Victoria  University,  Manchester. 
*PRIMER  OF  CHEMISTRY.     Illustrated.     With  Questions.     ISmo.     Is. 
♦LESSONS  IN  ELEMENTARY  CHEMISTRY,  INORGANIC  AND  ORGANIC. 
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Roscoe,  F.R.S.,  and  Prof.  C.  Sghorlemmer,  F.R.S.     Illustrated.    Svo. 
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larged by  the  Author  and  by  A.  Schuster,  F.R.S.,  Ph.D.,  Professor  of  Applied 
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*THORPE.— A  SERIES  OF  CHEMICAL  PROBLEMS.  With  Key.  For  use 
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ReAised  and  Enlarged  by  W.  Tate,  Assoc.N.S.S.  With  Preface  by  Sir  H.  E. 
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THORPE  and  RUGKER.— A  TREATISE  ON  CHEMICAL  PHYSICS.  By  Prof. 
T.  B.  Thorpe,  F.R.S,,  and  Prof.  A.  W.  RWcker,  F.R.S.    Illustrated.    Svo. 

[In  preparation. 

WILLIAMS.  — ELEMENTS  OF  CRYSTALLOGRAPHY  FOR  STUDENTS  OF 
CHEMISTRY,  PHYSICS,  AND  MINERALOGY.  By  G.  H.  Williams,  Ph.D. 
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FERREL,— A  POPULAR  TREATISE  ON  THE  WINDS.     Comprising  the  General 
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larged.   8vo.     12s. 
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48,  6d. 
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NATURAL  SCIENCES 


BIOLOO-Y. 

ALLEN.— ON  THE  COLOURS  OP  FLOWERS,  as  lUustrated  in  the  Britisli  Flora. 

By  Grant  Allen.  Illustrated.  Or.  8vo.  3s.  6d. 
BALFOUR.— A  TREATISE  ON  COMPARATIVE  EMBRYOLOGY.  By  F.  M. 
Balfour,  P.R.S.,  Fellow  and  Lecturer  of  Trinity  College,  Cambridge.  Illus- 
trated. 2d  Ed.,  reprinted  without  alteration  from  the  1st  Ed.  2  vols.  8vo. 
Vol.  L  18s.  Vol.  IL  21s. 
BALFOUR  and  WARD.— A  GENERAL  TEXT-BOOK  OF  BOTANY.  By  Isaac 
Bayley  Balfour,  F.R.S.,  Professorof  Botany  in  the  University  of  Edinburgh, 
and  H.  Marshall  Ward,  F.R.S.,  Professor  of  Botany  in  the  Royal  Indian 
Engineering  College,  Cooper's  HiU.    8vo.  [In  preparation 

*BETTANY.-FIRST  LESSONS  IN  PRACTICAL  BOTANY.     By  G.  T.  Bettanf. 

18mo.     Is. 
*BOWER.— A  COURSE  OF  PRACTICAL  INSTRUCTION  IN  BOTANY.    By  P. 
O.  Bower,  D.Sc,  Regius  Professor  of  Botany  in  the  University  of  Glasgow. 
Cr.  8vo.     10s.  6d. 
BUOKTON.  —monograph  OP  THE  BRITISH  CICADA,  OR  TETTIGID.®.    By 
G.  B.  BucKTON.     In  8  parts,  Quarterly.    Part  I.  January,  1890.     8vo.    Parts 
I.-V.  ready.    8s.  each,  net.    Vol  I.    33s.  6d.  net. 
CHURCH    and    SCOTT.  —  MANUAL    OF   VEGETABLE    PHYSIOLOGY.      By 
Professor  A.  H.  Church,  and  D.  H.  Scott,  D.Sc,  Lecturer  in  the  Normal 
School  of  Science.     Illustrated.     Cr.  8vo.  [In  preparation. 

COPE.— THE  ORIGIN  OF  THE  FITTEST.     Essays  on  Evolution.     By  E.   D. 

Cope,  M.A.,  Ph.D.     8vo.     12s.  6d. 
COUES.— HANDBOOK    OF    FIELD  AND    GENERAL   ORNITHOLOGY.     By 

Prof.  Elliott  Coues,  M.A.     Illustrated.    8vo.     10s.  net. 
DARWIN.— MEMORIAL  NOTICES  OF  CHARLES  DARWIN,  F.R.S.,  etc.     By 
T.  H.  Huxley,  F.R.S.,  G.  J.  Romanes,  F.R.S.,  Archibald  Geikie,  F.R.S., 
and  W.  Thiselton  Dyer,  F.R.S.     Reprinted  from  Nature.     With  a  Por- 
trait.    Cr.  8vo.     2s.  6d. 
EIMER.— ORGANIC  EVOLUTION  AS  THE  RESULT  OF  THE  INHERITANCE 
OF  ACQUIRED  CHARACTERS  ACCORDING  TO  THE  LAWS  OF  OR- 
GANIC GROWTH.     By  Dr.  G.  H.  Theodor  Eimer.     Translated  by  J.  T. 
Cunningham,  F.R.S.E.,  late  Fellow  of  University  College,  Oxford.  8vo.  12s.  6d. 
FEARNLEY.— A   MANUAL  OF    ELEMENTARY  PRACTICAL   HISTOLOGY. 

By  William  Fearnley.    Illustrated.    Cr.  8vo.    7s.  6d. 
FLOWER  and  GADOW.— AN  INTRODUCTION  TO  THE  OSTEOLOGY  OF 
THE  MAMMALIA.     By  W.  H.  Flower,  F.R.S.,  Director  of  the  Natural  His- 
tory Departments  of  the  British  Museum.    Illustrated.    8d  Ed.    Revised  with 
the  assistance  of  Hans  Gadow,  Ph.D.,  Lecturer  on  the  Advanced  Morphology 
of  Vertebrates  in  the  University  of  Cambridge.    Cr,  8vo.    10s.  6d. 
FOSTER.— Works  by  Michael  Foster,  M.D.,  Professor  of  Physiology  in  the  Uni- 
versity of  Cambridge. 
*PRIMER  OF  PHYSIOLOGY.    lUustrated.    18mo.     Is. 

A  TEXT-BOOK  OF  PHYSIOLOGY.  Hlustrated.  6th  Ed.,  largely  revised.  8vo. 
Part  I.,  comprising  Book  I.  Blood— The  Tissues  of  Movement,  The  Vascular 
Mechanism.  10s.  6d.  Part  IL,  comprising  Book  II.  The  Tissues  of  Chemical 
Action,  with  their  Respective  Mechanisms— Nutrition.  10s.  6d.  Part  III. 
The  Central  Nervous  System.  7s.  6d. 
OSTER  and  BALFOUR.  — THE  ELEMENTS  OF  EMBRYOLOGY.  By  Prof. 
Michael  Foster,  M.D.,  and  the  late  F.  M.  Balfour,  F.R.S.,  Professor  of 
Animal  Morphology  in  the  University  of  Cambridge.  2d  Ed.,  revised.  Edited 
by  A.  Sedgwick,  M.A.,  Fellow  and  Assistant  Lecturer  of  Trinity  CoUege, 
Cambridge,  and  W.  Heape,  M.A.,  late  Demonstrator  in  the  Morphological 
Laboratory  of  the  University  of  Cambridge.  Illustrated.  Cr.  8vo.  10s.  6d. 
FOSTER  and  LANGLEY.— A  COURSE  OF  ELEMENTARY  PRACTICAL 
PHYSIOLOGY  AND  HISTOLOGY.  By  Prof.  Michael  Foster,  M.D.,  and 
J.  N.  Langley,  F.R.S.,  Fellow  of  Trinity  College,  Cambridge.  6th  Ed.  Cr. 
8vo.    7s.  6d. 


BIOLOGY  37 

GAMGEE.— A  TEXT -BOOK  OF  THE  PHYSIOLOGICAL  CHEMISTRY  OF 
THE  ANIMAL  BODY.  Including  an  Account  of  the  Chemical  Changes 
occurring  in  Disease.  By  A.  Gamgee,  M.D.,  F.R.S.  Illustrated.  8vo.  Vol. 
L     18s. 

GOODALE.— PHYSIOLOGICAL  BOTANY.  I.  Outlines  of  the  Histology  of 
Phsenoganious  Plants.  II.  Vegetable  Physiology.  By  George  Lincoln 
GooDALE,  M.A.,  M.D.,  Professor  of  Botany  in  Harvard  University.  Svo. 
,10s.  6d. 

GRAY.— STRUCTURAL  BOTANY,  OR  ORGANOGRAPHY  ON  THE  BASIS 
OP  MORPHOLOGY.  To  which  are  added  the  Principles  of  Taxonomy  and 
Phytogi'aphy,  and  a  Glossary  of  Botanical  Terms.  By  Prof.  Asa  Gray,  LL.D. 
Svo.  10s.  6d. 
THE  SCIENTIFIC  PAPERS  OF  ASA  GRAY.  Selected  by  C.  Sprague  Sar- 
gent. 2  vols.  Vol.  I.  Reviews  of  Works  on  Botany  and  Related  Subjects, 
1834-1887.    Vol.  IL    Essays,  Biographical  Sketches,  1841-1886.    Svo.    21s. 

HAMILTON.— A  SYSTEMATIC  AND  PRACTICAL  TEXT-BOOK  OF  PATHO- 
LOGY. By  D.  J.  Hamilton,  F.R.S.B.,  Professor  of  Pathological  Anatomy 
in  the  University  of  Aberdeen.    Illustrated.    Svo.    Vol.  I.     25s. 

HARTIG.— TEXT -BOOK  OF  THE  DISEASES  OF  TREES.  By  Dr.  Robert 
Hartio.  Translated  by  Wm.  Somerville,  B.Sc,  D.CE.,  Lecturer  on  Forestry 
in  the  University  of  Edinburgh.  Edited,  with  Introduction,  by  Prof.  H. 
Marshall  Ward.    Svo.  [In  preparation. 

HGOKIiR.- Works  by  Sir  Joseph  Hooker,  F.R.S.,  &c. 
*PRIMER  OF  BOTANY.    Illustrated.     ISmo.     Is. 

THE  STUDENT'S  FLORA  OF  THE  BRITISH  ISLANDS.      3d  Ed.,  revised. 
GI.  Svo.     10s.  6d. 

HOWES.— AN  ATLAS  OF  PRACTICAL  ELEMENTARY  BIOLOGY.  By  G.  B. 
Howes,  Assistant  Professor  of  Zoology,  Normal  School  of  Science  and  Royal 
School  of  Mines.    With  a  Preface  by  Prof.  T.  H.  Huxley,  F.R.S.    4to.     14s. 

HUXLEY.— Works  by  Prof.  T.  H.  Huxley,  F.R.S. 
^INTRODUCTORY  PRIMER  OF  SCIENCE.     ISmo.    Is. 

^LESSONS  IN  ELEMENTARY  PHYSIOLOGY.    Illustrated.    Fcap.  Svo.  4s.  6d. 
*QUESTIONS  ON  HUXLEY'S  PHYSIOLOGY.     By  T.  Alcock,  M.D.     ISmo. 
Is.  6d. 

HUXLEY  and  MARTIN.— A  COURSE  OF  PRACTICAL  INSTRUCTION  IN 
ELEMENTARY  BIOLOGY.  By  Prof.  T.  H.  Huxley,  F.R.S.,  assisted  by 
H.  N.  Martin,  F.R.S.,  Professor  of  Biology  in  the  Johns  Hopkins  University, 
U.S.A.  New  Ed.,  revised  and  extended  by  G.  B.  Howes  and  D.  H.  Scott, 
Ph.D.,  Assistant  Professors,  Normal  School  of  Science  and  Royal  School  of 
Mines,     With  a  Preface  by  T.  H.  Huxley,  F.R.S.    Cr.  Svo.     10s.  6d. 

KLEIN.— Works  by  E.  Klein,  F.R.S.,  Lecturer  on  General  Anatomy  and  Physio- 
logy in  the  Medical   School  of  St.   Bartholomew's  Hospital,  Professor  of 
Bacteriology  at  the  College  of  State  Medicine,  London. 
MICRO-ORGANISMS  AND  DISEASE.     An  Introduction  into  the  Study  of 

Specific  Micro-Organisms.    Illustrated.    3d  Ed.,  revised.    Cr.  Svo.     6s. 
THE  BACTERIA  IN  ASIATIC  CHOLERA.     Cr.  Svo.     6s. 

LANG.— TEXT-BOOK  OF  COMPARATIVE  ANATOMY.  By  Dr.  Arnold  Lang, 
Professor  of  Zoology  in  the  University  of  Zurich.  Translated  by  H.  M.  Bernard, 
M.A.,  and  M.  Bernard.  Introduction  by  Prof.  E.  Haeckel.  2  vols.  Illus- 
trated.   Svo.  [In  the  Press. 

LANKESTER.— Works  by  E.   Ray   Lankester,    F.R.S.,   Linacre  Professor    of 
Human  and  Comparative  Anatomy  in  the  University  of  Oxford. 
A  TEXT-BOOK  OP  ZOOLOGY.    Svo.  [In  preparation. 

THE  ADVANCEMENT  OF  SCIENCE.    Occasional  Essays  and  Addresses.  Svo. 
10s.  6d. 

LUBBOCK.— Works  by  the  Right  Hon.  Sir  John  Lubbock,  F.R.S,,  D.O.L. 
THE  ORIGIN  AND  METAMORPHOSES  OF  INSECTS.  Illustrated.   Cr.  Svo. 
33.  6d. 


38  NATURAL  SCIENCES 

ON  BRITISH   WILD   FLOWERS    CONSIDERED   IN  RELATION    TO    IN- 
SECTS.    Illustrated.     Cr.  8vo.    4s.  6d. 
FLOWERS,  FRUITS,  AND  LEAVES.    Illustrated.     2d  Ed.    Cr.  8vo.    4s.  6d. 
SCIENTIFIC  LECTURES.     2d  Ed.     8vo.     8s.  6d. 
FIFTY  YEARS  OF  SCIENCE.     Being  the  Address  delivered  at  York  to  the 

British  Association,  August  1881.    5th  Ed.    Cr.  8vo.    2s.  6d. 
MARTIN  and  MOALE.-ON  THE  DISSECTION  OF  VERTEBRATE  ANIMALS. 

By  Prof.  H.  N.  Maetin  and  W.  A.  Moale.    Cr.  8vo.  [In  preparation. 

MIVART.— LESSONS  IN  ELEMENTARY  ANATOMY.    By  St.  George  Mivart 

F.R.S.,  Lecturer  on  Comparative  Anatomy  at  St.  Mary's  Hospital.    Illustrated, 

Fcap.  8vo.     6s.  6d. 
MtfLLER.— THE   FERTILISATION  OF   FLOWERS.      By  Hermann  Miller, 

Translated  and  Edited  by  D'Arcy  W.  Thompson,  B.A.,  Professor  of  Biology  in 

University  College,  Dundee.     With  a  Preface  by  C.  Darwin,  F.R.S.     Illus 

trated.     8vo.     21s. 
OLIVER.— Works  by  Daniel  Oliver,  F.R.S.,  late  Professor  of  Botany  in  Uni 

versity  College,  London. 
*LESSONS  IN  ELEMENTARY  BOTANY.    Illustrated.     Fcap.  8vo.    4s.  6d. 
FIRST  BOOK  OF  INDIAN  BOTANY.     Illustrated.    Ex.  fcap.  Svo.    6s.  6d. 
PARKER.— Works  by  T.  Jeffery  Parker,  F.R.S.,  Professor  of  Biology  in  the 

University  of  Otago,  New  Zealand. 
A  COURSE  OP  INSTRUCTION  IN  ZOOTOMY  (VERTEBRATA).   Illustrated. 

Cr.  Svo.    8s.  6d. 
LESSONS  IN  ELEMENTARY  BIOLOGY.    Illustrated.    Cr.  Svo.    [In  tU  Press. 
PARKER  and  BETTANY.— THE  MORPHOLOGY  OF  THE  SKULL.    By  Prof. 

W.  K.  Parker,  F.R.S.,  and  G.  T.  Bettany.     Illustrated.     Cr.  Svo.     10s.  6d. 
ROMANES.— THE    SCIENTIFIC    EVIDENCES    OF   ORGANIC    EVOLUTION. 

By  George  J.  Romanes,  F.R.S.,  Zoological  Secretary  of  the  Linnean  Society. 

Cr.  Svo.     2s.  6d. 
SEDGWICK.- A  SUPPLEMENT  TO  F.  M.  BALFOUR'S  TREATISE  ON  EM- 
BRYOLOGY.    By  Adam  Sedgwick,  F.R.S.,  Fellow  and  Lecturer  of  Trinity 

College,  Cambridge.     Illustrated.     Svo.  [In  preparation. 

SHUFELDT.— THE  MYOLOGY  OF  THE  RAVEN  {Corvus  corax  smuatus).     A 

Guide  to  the  Study  of  the  Muscular  System  in  Birds.    By  R.  W.  Shtjfeldt. 

Illustrated.     Svo.     13s.  net. 
SMITH.— DISEASES    OF   FIELD  AND   GARDEN  CROPS,   CHIEFLY  SUCH 

AS  ARE  CAUSED  BY  FUNGI.    By  W.  G.  Smith,  F.L.S.    Illustrated.     Fcap. 

Svo.    4s.  6d. 
STEWART  and  CORRY.— A  FLORA  OF  THE  NORTH-EAST  OF  IRELAND. 

Including  the  Phanerogamia,  the  Cryptogamia  Vascularia,  and  the  Muscinese. 

By  S.  A.  Stewart,  Curator  of  the  Collections  in  the  Belfast  Museum,  and  the 

late  T.  H.  Corry,  M.A.,  Lecturer  on  Botany  in  the  University  Medical  and 

Science  Schools,  Cambridge.     Cr.  Svo.     5s.  6d. 
WALLACE.— DARWINISM  :  An  Exposition  of  the  Theory  of  Natural  Selection, 

with  some  of  its  Applications.    By  Alfred  Russel  Wallace,  LL.D.,  F.R.S. 

3d  Ed.     Cr.  Svo.    9s. 
WARD.— TIMBER  AND  SOME  OF  ITS  DISEASES.     By  H.  Marshall  Ward, 

F.R.S.,  Professor  of  Botany  in  the  Royal  Indian  Engineering  College,  Cooper's 

Hill.     Illustrated.     Cr.  Svo.     6s. 
WIEDERSHEIM.— ELEMENTS    OP    THE     COMPARATIVE    ANATOMY    OF 

VERTEBRATES.     By  Prof.   R.   Wiedersheim.     Adapted   by  W.   Newton 

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MEDICINE. 

BLYTH.— A  MANUAL  OF  PUBLIC  HEALTH.     By  A.  Wynter  Blyth,  M.R.C.S. 
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MEDICINE  39 

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Royal  College  of  Physicians,  London, 
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Museum.    Illustrated.    Cheaper  Issue.    Svo,    5s. 
ON    THE    CONNECTION   BETWEEN   CHEMICAL    CONSTITUTION  AND 
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GRIFFITHS.— LESSONS  ON  PRESCRIPTIONS  AND  THE  ART  OF  PRE- 
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Water— Cold  Water  as  an  Antipyretic— Artificial  Baths— On  Lavage,  or 
washing  out  the  Stomach— Massage — The  Weir  Mitchell  Method— Venesection 
— Electricity,  Physical  and  Physiological  Facts  of  Importance  in  Medicine— 
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By  Prof.  J.  M.  Baldwin,  M.A.,  LL.D.     2d  Ed.,  revised.    Svo.     123.  6d. 
BOOLE.— THE  MATHEMATICAL  ANALYSIS  OF  LOGIC.      Being  an  Essay 

towards  a  Calculus  of  Deductive  Reasoning.    By  George  Boole.    Svo.    6s. 
OALDERWOOD.— HANDBOOK  OF  MORAL  PHILOSOPHY.    By  Rev.  Henry 

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Edinburgh.    14th  Ed.,  largely  rewritten.    Or.  Svo.    6s. 


40  HUMAN  SCIENCES 

CLIFFORD.— SEEING  AND  THINKING.     By  the  late  Prof.  W.  K.  Clifford, 

F.R.S.    With  Diagrams.     Or.  8vo.    33.  6d. 
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*PRIMER  OF  LOGIC.    18mo.    Is. 

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POLITICAL  ECONOMY  41 

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PANTALEONI.— MANUAL  OF  POLITICAL  ECONOMY.  By  Prof.  M.  Panta- 
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SIDGWICK.— THE  PRINCIPLES  OF  POLITICAL  ECONOMY.  By  Henry 
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42  HUMAN  SCIENCES 


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ADAMS  and  CUNNINGHAM.— THE  SWISS  CONFEDERATION.    By  Sir  F.  O, 

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Cr.  8vo.    2s.  6d. 
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CONQUEST.    The  Norman  Period,  1066-1204.     By  Melville  M.  Bigelow, 
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BOUTMY.  —  STUDIES    IN    CONSTITUTIONAL    LAW.      By    Emile    BouxMy. 
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*BUCKLAND.— OUR  NATIONAL  INSTITUTIONS.    A  Short  Sketch  for  Schools. 

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CHERRY.— LECTURES  ON  THE  GROWTH  OF  CRIMINAL  LAW  IN  ANCIENT 
COMMUNITIES.     By  R.  R.  Cherry,  LL.D.,  Reid  Professor  of  Constitutional 
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DICEY.— INTRODUCTION  TO  THE  STUDY  OF  THE  LAW  OF  THE  CONSTITU- 
TION.   By  A.  V.  Dicey,  B.C.L.,  Vinerian  Professor  of  English  Law  in  the 
University  of  Oxford.     3d  Ed.    8vo.    12s.  6d. 
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ANTHROPOLOGY^ — EDUCATION  43 

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ITINERANT,  IN  THE  FIFTH  YEAR  OF  THE  REIGN  OP  KING  HENRY 

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ANTHROPOLOGY. 

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G.  Prazer,  M.A.,  Fellow  of  Trinity  College,  Cambridge.     2  vols.     Svo.     28s. 

M'LENNAN.— THE  PATRIARCHAL  THEORY.  Based  on  the  papers  of  the  late 
John  P.  M'Lennan.  Edited  by  Donald  M'Lennan,  M.A.,  Barrister-at-Law. 
Svo.  14s. 
STUDIES  IN  ANCIENT  HISTORY.  Comprising  a  Reprint  of  "Primitive 
Marriage."  An  inquiry  into  the  origin  of  the  form  of  capture  in  Marriage 
Ceremonies.    Svo.    16s. 

TYLOR.— ANTHROPOLOGY.    An  Introduction  to  the  Study  of  Man  and  Civilisa- 
tion.   By  E.  B.  Tylor,  F.R.S.     Illustrated.    Cr.  Svo.    7s.  6d. 

WESTERMARGK.— THE  HISTORY  OF  HUMAN  MARRIAGE.    By  Dr  Edward 
Westermarck.  [In  preparation. 

EDUCATION. 

ARNOLD.-  REPORTS  ON  ELEMENTARY  SCHOOLS.    1852-1882.   By  Matthew 
Arnold,  D.C.L.    Edited  by  the  Right  Hon.  Sir  Francis  Sandford,  K.C.B. 
Cheaper  Issue.    Cr.  Svo.     3s.  6d. 
HIGHER  SCHOOLS  AND  UNIVERSITIES  IN  GERMANY.     By  the  same. 
Crown  Svo.    6s. 


44  TECHNICAL  KNOWLEDGE 

BALL.— THE  STUDENT'S  GUIDE  TO  THE  BAR.    By  Walter  W.  R.  Ball, 

M.A.,  Fellow  and  Assistant  Tutor  of  Trinity  College,  Cambridge.    4th  Ed., 

revised.    Cr.  8vo.    2s.  6d. 
*BLAKISTON.— THE  TEACHER.     Hints  on  School  Management.    A  Handbook 

for  Managers,  Teachers'  Assistants,  and  Pupil  Teachers.    By  J.  R.  Blakiston. 

Or.  8vo.    2s.  6d.    (Recommended  by  the  London,  Birmingham,  and  Leicester 

School  Boards.) 
CALDERWOOD.— ON  TEACHING.    By  Prof.  Henry  Calderwood.    New  Ed. 

Ex.  fcap.  8vo.     2s.  6d. 
FEARON.-SCHOOL  INSPECTION.  By  D.  R.  Fearon.    6th  Ed.   Cr.  Svo.   2s.  6d. 
FITCH.— NOTES   ON   AMERICAN   SCHOOLS  AND  TRAINING   COLLEGES. 

Reprinted  from  the  Report  of  the  English  Education  Department  for  1888-89, 

with  permission  of  the  Controller  of  H.M.'s  Stationery  Office.     By  J.   G. 

Fitch,  M.A.    G1.  Svo.     2s.  6d. 
GEIKIE.— THE  TEACHING  OF  GEOGRAPHY.    A  Practical  Handbook  for  the 

use  of  Teachers.     By  Archibald  Geikie,  F.R.S.,  Director -General  of  the 

Geological  Survey  of  the  United  Kingdom.     Cr.  Svo.     2s. 
GLADSTONE.— SPELLING  REFORM  FROM  A  NATIONAL  POINT  OF  VIEW. 

By  J.  H.  Gladstone.     Cr.  Sa'o.     Is.  6d. 
HERTEL.— OVERPRESSURE  IN  HIGH  SCHOOLS  IN  DENMARK.      By  Dr. 

Hertel.     Translated  by  C.   G.   Sorensen.      With  Introduction  by  Sir  J. 

Crichton-Browne,  F.R.S.    Cr.  Svo.     3s.  6d. 
TODHUNTER.— THE  CONFLICT  OF  STUDIES.     By  Isaac  Todhunter,  F.R.S. 

Svo.    10s.  6d. 

TECHNICAL  KNOWLEDGE. 

(See  also  MECHANICS,  LAW,   and  MEDICINE.) 

Civil  and  Mechanical  Engineering;  Military  and  Naval  Science; 

Agriciiltiire ;  Domestic  Economy ;  Book-Keeping. 

CIVIL  AND  MECHANICAL  ENGINEERING. 

ALEXANDER  and  THOMSON.— ELEMENTARY  APPLIED  MECHANICS.  By 
T.  Alexander,  Professor  of  Civil  Engineering,  Trinity  College,  Dublin,  and 
A.  W.  Thomson,  Professor  at  College  of  Science,  Poona,  India.  Part  II. 
Transverse  Stress.     Cr.  8vo.     10s.  6d. 

CHALMERS.  -GRAPHICAL  DETERMINATION  OF  FORCES  IN  ENGINEER- 
ING STRUCTURES.     By  J.  B.  Chalmers,  C.E.     Illustrated.     8vo.     24s. 

COTTERILL.— APPLIED  MECHANICS :  An  Elementary  General  Introduction  to 
the  Theory  of  Structures  and  Machines.  By  J.  H.  Cottertll,  F.R.S.,  Pro- 
fessor  of  Applied  Mechanics  in  the  Royal  Naval  College,  Greenvi^ich.  2d  Ed. 
8vo.     18s. 

COTTERILL  and  SLADE.— LESSONS  IN  APPLIED  MECHANICS.  By  Prof. 
J.  H.  Cotterill  and  J.  H.  Slade.     Fcap.  8vo.    5s.  6d. 

GRAHAM.  —  GEOMETRY  OF  POSITION.  By  R.  H.  Graham.  Ct.  Svo. 
7s.  6d. 

KENNEDY.— THE  MECHANICS  OF  MACHINERY.  By  A.  B.  W.  Kennedy, 
F.R.S.     Illustrated.    Cr.  Svo.    12s.  6d. 

WHITHAM.— STEAM-ENGINE  DESIGN.  For  the  Use  of  Mechanical  Engineers, 
Students,  and  Draughtsmen.  By  J.  M.  Whitham,  Professor  of  Engineering, 
Arkansas  Industrial  University.     Illustrated.    Svo.     25s. 

YOUNG.— SIMPLE  PRACTICAL  METHODS  OF  CALCULATING  STRAINS 
ON  GIRDERS,  ARCHES,  AND  TRUSSES.  With  a  Supplementa,ry  Essay  on 
Economy  in  Suspension  Bridges.  By  B.  W.  Young,  C.E.  With  Diagrams. 
Svo.    7s.  6d. 

MILITARY   AND   NAVAL   SCIENCE. 

AITKEN.— THE  GROWTH  OF  THE  RECRUIT  AND  YOUNG  SOLDIER.  With 
a  view  to  the  selection  of  "Growing  Lads"  for  the  Army,  and  a  Regulated 
System  of  Training  for  Recruits.  By  Sir  W.  Aitken,  F.R.S.,  Professor  of 
Pathology  in  the  Army  Medical  School.    Cr.  Svo.    8s.  6d. 


MILITARY  SCIENCE — AGRICULTURE  45 

ARMY  PRELEMINARY  EXAMINATION,  1882-1889,  Specimens  of  Papers  set  at 
the.  With  Answers  to  the  Mathematical  Questions.  Subjects :  Arithmetic, 
Algebra,  Euclid,  Geometrical  Drawing,  Geography,  French,  English  Dictation. 
Cr.  8vo.     33.  6d. 

MATTHEWS.— MANUAL  OF  LOGARITHMS.    By  G.  F.  Matthews,  B.A.    8vo. 

MERCUR.— ELEMENTS  OP  THE  ART  OF  WAR.  Prepared  for  the  use  of 
Cadets  of  the  United  States  Military  Academy.  By  James  Mercur,  Professor 
of  Civil  Engineering  at  the  United  States  Academy,  West  Point,  New  York, 
2d  Ed.,  revised  and  corrected.    8vo.    17s. 

PALMER.— TEXT -BOOK  OF  PRACTICAL  LOGARITHMS  AND  TRIGONO- 
METRY. By  J.  H.  Palmer,  Head  Schoolmaster,  R.N.,  H.M.S.  Cambridge, 
Devonport.     Gl.  Svo.    4s.  6d. 

ROBINSON.— TREATISE  ON  MARINE  SURVEYING.  Prepared  for  the  use  of 
younger  Naval  Officers.  With  Questions  for  Examinations  and  Exercises 
principally  from  the  Papers  of  the  Royal  Naval  College.  With  the  results. 
By  Rev.  John  L.  Robinson,  Chaplain  and  Instructor  in  the  Royal  Naval 
College,  Greenwich.     Illustrated.    Cr.  Svo.     Vs.  6d. 

SANDHURST  MATHEMATICAL  PAPERS,  for  Admission  into  the  Royal  Military 
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matics at  the  Royal  Military  Academy,  Woolwich.    Cr.  8vo.    3s.  6d. 

SHORTLAND.— NAUTICAL  SURVEYING.  By  the  late  Vice- Admiral  Shortland, 
LL.D.    Svo.    2l8. 

THOMSON.— POPULAR  LECTURES  AND  ADDRESSES.  By  Sir  William 
Thomson,  LL.D.,  P.R.S.  In  8  vols.  Illustrated.  Cr.  Svo.  Vol.  III.  Papers 
on  Navigation.  [In  the  Press. 

WILELINSON.— THE  BRAIN  OP  AN  ARMY.  A  Popular  Account  of  the  German 
General  Staff.     By  Spenser  Wilkinson.    Cr.  Svo.     2s.  6d. 

WOLSELEY.— Works  by  General  Viscount  Wolseley,  G.C.M.G. 
THE  SOLDIER'S  POCKET-BOOK  FOR  FIELD  SERVICE.     6th  Ed.,  revised 

and  enlarged.     16mo.     Roan.     5s. 
FIELD  POCKET-BOOK  FOR  THE  AUXILIARY  FORCES.     16mo.     Is.  6d. 

WOOLWICH  MATHEMATICAL  PAPERS,  for  Admission  into  the  Royal  Military 
Academy,  Woolwich,  1S80-18SS  inclusive.  Edited  by  E.  J.  Brooksmith,  B.A., 
Instructor  in  Mathematics  at  the  Royal  Military  Academy,  Woolwich.  Cr. 
Svo.    6s. 

AGRICULTURE. 

FRANKLAND.— AGRICULTURAL  CHEMICAL  ANALYSIS,  A  Handbook  of. 
By  Percy  F.  Frankland,  F.R.S.,  Professor  of  Chemistry,  University  College, 
Dundee.  Founded  upon  Leitfaden  fur  die  Agriculture  Chemiche  Analyse,  von 
Dr.  F.  Krooker.  Cr.  Svo.  7s.  6d. 
HARTIG.— TEXT -BOOK  OF  THE  DISEASES  OF  TREES.  By  Dr.  Robert 
Hartig.  Translated  by  Wm.  Somerville,  B.Sc,  D.CE.,  Lecturer  on  Forestry 
in  the  University  of  Edinburgh.  Edited,  with  Introduction,  by  Prof.  H. 
Marshall  Ward.     Svo.  [In  preparation. 

LASLETT.— TIMBER  AND  TIMBER  TREES,  NATIVE  AND  FOREIGN.      By 

Thomas  Laslett.     Cr.  Svo.    8s.  6d. 
SMITH.— DISEASES  OF  FIELD  AND  GARDEN  CROPS,  CHIEFLY  SUCH  AS 
ARE  CAUSED  BY  FUNGI.    By  Worthington  G.  Smith,  F.L.S.   Illustrated. 
Fcap.  Svo.     4s.  6d. 
TANNER.— *ELEMBNTARY  LESSONS  IN  THE  SCIENCE  OF  AGRICULTURAL 
PRACTICE.    By  Henry  Tanner,  F.C.S.,  M.R.A.C,  Examiner  in  the  Prin- 
ciples of  Agriculture  under  the  Government  Department  of  Science.    Fcap. 
Svo.    3s.  6d. 
*FIRST  PRINCIPLES  OF  AGRICULTURE.    By  the  same.     ISmo.     Is. 
THE  PRINCIPLES  OF  AGRICULTURE.     By  the  same.    A  Series  of  Reading 
Books  for  use  in  Elementary  Schools.    Ex.  fcap.  8v% 
*I.  The  Alphabet  of  the  Principles  of  Agriculture.     6d. 
*II.  Further  Steps  in  the  Principles  of  .^riculture.     Is. 
*III.  Elementary  School  Readings  on  the  Principles  of  Agriculture  for  the 
third  stage.     Is. 


46  TECHNICAL  KNOWLEDGE 

WARD.— TIMBER  AND  SOME  OF  ITS  DISEASES.  By  H.  Marshall  Ward, 
M.A.,  F.L.S.,  P.R.S.,  Fellow  of  Christ's  College,  Cambridge,  Professor  of 
Botany  at  the  Royal  Indian  Engineering  College,  Cooper's  Hill.  With  Illustra- 
tions.   Cr.  8vo.     6s. 

DOMESTIC  ECONOMY. 

*BARKER.— FIRST  LESSONS  IN  THE  PRINCIPLES  OF  COOKING.    By  Lady 

Barker.    18mo.     Is. 
*BERNERS,— FIRST  LESSONS  ON  HEALTH.     By  J.  Berners.    18mo.    Is. 
»COOKERY  BOOK.— THE  MIDDLE  CLASS  COOKERY  BOOK.     Edited  by  the 

Manchester  School  of  Domestic  Cookery.     Fcap.  8vo.     Is.  6d. 
CRAVEN.- A  GUIDE  TO  DISTRICT  NURSES.     By  Mrs.  Dacre  Craven  {nh 

Florence  Sarah  Lees),  Hon.  Associate  of  the  Order  of  St.  John  of  Jerusalem, 

etc.     Cr..8vo.     2s.  6d. 
FREDERICK.— HINTS    TO    HOUSEWIVES    ON    SEVERAL    POINTS,    PAR- 

TICULARLY  ON  THE  PREPARATION  OF  ECONOMICAL  AND  TASTEFUL 

DISHES.     By  Mrs.  Frederick,     Cr.  8vo.     Is. 
*GRAND'HOMME.— CUTTING-OUT  AND  DRESSMAKING.    From  the  French  of 

Mdlle.  E.  Grand'homme.     With  Diagrams.     18mo.    Is. 
JEX-BLAKE.— THE  CARE  OF  INFANTS.    A  Manual  for  Mothers  and  Nurses. 

By  Sophia  Jex-Blake,  M.D.,  Lecturer  on  Hygiene  at  the  London  School  of 

Medicine  for  Women.    18mo.    Is. 
RATHBONE.— THE   HISTORY  AND   PROGRESS  OP   DISTRICT  NURSING 

FROM   ITS   COMMENCEMENT  IN  THE  YEAR  1859  TO  THE  PRESENT 

DATE,  including  the  foundation  by  the  Queen  of  the  Queen  Victoria  Jubilee 

Institute  for  Nursing  the  Poor  in  their  own  Homes.    By  William  Rathbone, 

M.P.     Cr.  8vo.     23.  6d. 
*TEGETMEIER.— HOUSEHOLD  MANAGEMENT  AND  COOKERY.     With  an 

Appendix  of  Recipes  used  by  the  Teachers  of  the  National  School  of  Cookery. 

By  W.  B.  Tegetmeier,    Compiled  at  the  request  of  the  School  Board  for 

London.    18mo.    Is. 
»WRIGHT.— THE  SCHOOL  COOKERY-BOOK.    Compiled  and  Edited  by  C.  B. 

Guthrie  Wright,  Hon.  Sec.  to  the  Edinburgh  School  of  Cookery.    18mo.    Is. 

BOOK-KEEPING-. 

*THORNTON.— FIRST    LESSONS  IN    BOOK-KEEPING.      By    J.   Thornton. 

Cr.  8vo.     2s.  6d.    KEY.     Oblong  4to.     10s.  6d. 
*PRIMER  OF  BOOK-KEEPING.     By  the  same.     18mo.    Is. 
KEY.     Svo.     ^s.  6d. 

GEOGRAPHY. 

(See  also  PHYSICAL  GEOGRAPHY.) 

BARTHOLOMEW.— *THB  ELEMENTARY  SCHOOL  ATLAS.     By  John  Bar- 
tholomew,  F.R.G.S.     4to.     Is. 
*MACMILLAN'S  SCHOOL  ATLAS,  PHYSICAL  AND  POLITICAL.     Consisting 
of  80  Maps  and  complete  Index.    By  the  same.     Prepared    for  the  use  of 
Senior  Pupils.     Royal  4to.     8s.  6d.     Half-morocco.    10s.  6d. 
THE   LIBRARY    REFERENCE  ATLAS  OF  THE    WORLD.     By  the  same. 
A  Complete  Series  of  84  Modern  Maps.    With  Geographical  Index  to  100,000 
places.     Half- morocco.    Gilt  edges.     Folio.     £2: 12: 6  net.     Also  issued  in 
parts,  5s.  each  net.     Geographical  Index,  7s.  6d.  net.     Part  I.,  April  1891. 
*CLARKE.— CLASS-BOOK  OP  GEOGRAPHY.     By  C.  B.  Clarke,  F.R.S.    New 

Ed.,  revised  1889,  with  18  Maps.     Fcap.  Svo.     3s.     Sewed,  2s.  6d. 
GEIKIE.— Works  by  Archibald  Geikie,  F.R.S.,  Director-General  of  the  Geological 

Survey  of  the  United  Kingdom. 
*TnE  TEACHING  OF  GEOGRAPHY.     A  Practical  Handbook  for  the  use  of 

Teachers.     Cr.  8vo.     2s. 
^GEOGRAPHY  OP  THE  BRITISH  ISLES.     18mo.     Is. 


GEOGRAPHY — HISTORY  47 

*GREEN.— A  SHORT  GEOGRAPHY  OP  THE  BRITISH  ISLANDS.     By  John 

Richard  Green  and  A.  S.  Green.     With  Maps,     Fcap.  8vo.    3s.  6d. 
*GROVE.— A  PRIMER  OP  GEOGRAPHY.     By  Sir  George  Grove,  D.C.L. 

Illustrated.    ISmo.     Is. 
KIEPERT.— A  MANUAL  OF  ANCIENT  GEOGRAPHY.     By  Dr.  H.  Kiepert. 

Or.  8vo.     5s. 
MAOMILLAN'S   GEOGRAPHICAL    SERIES.  —  Edited  by  Archibald   Geikie, 

P.R.S.,  Director-General  of  the  Geological  Survey  of  the  United  Kingdom. 
*THE  TEACHING  OP  GEOGRAPHY.    A  Practical  Handbook  for  the  Use  of 

Teachers.    By  Archibald  Geikie,  F.R.S.    Cr.  8vo.    2s. 
*MAPS  AND  MAP-DRAWING.     By  W.  A.  Elderton.    ISmo.    Is. 
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*AN  ELEMENTARY  CLASS-BOOK  OF  GENERAL  GEOGRAPHY.    By  H.  R. 
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the  Heriot-Watt  College,  Edinburgh.    Illustrated.     Cr.  8vo.     3s.  6d. 
♦GEOGRAPHY  OP  EUROPE.     By  J.  Sime,  M.A.     Illustrated.    Gl.  8vo.    Ss. 
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F.  Blanford,  F.G.S.    Gl.  8vo.     2s.  6d. 
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Sutherland.  [In  preparation. 

*4t*  Other  volumes  will  be  announced  in  due  course. 

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*TOZER.— A  PRIMER  OF  CLASSICAL  GEOGRAPHY.    By  H.  P.  Tozer,  M.A. 

18mo.    Is. 

HISTORY. 

ARNOLD.— THE  SECOND  PUNIC  WAR.    Being  Chapters  from  THE  HISTORY 

OF  ROME,  by  the  late  Thomas  Arnold,  D.D.  ,  Headmaster  of  Rugby.    Edited, 

with  Notes,  by  W.  T.  Arnold,  M.A.     With  8  Maps.     Cr.  8vo.    5s. 
ARNOLD.— THE  ROMAN  SYSTEM  OP  PROVINCIAL  ADMINISTRATION  TO 

THE  ACCESSION  OP  CONSTANTINE  THE  GREAT.    By  W.  T.  Arnold, 

M.A.    Cr.  8vo.    6s. 
*BEESLY.— STORIES  FROM  THE  HISTORY  OF  ROME.     By  Mrs.  Beeslt. 

Fcap.  8vo.    2s.  6d. 
BRTOE.— Works  by  James  Bryce,  M.P.,  D.C.L.,  Regius  Professor  of  Civil  Law 

in  the  University  of  Oxford. 
THE  HOLY  ROMAN  EMPIRE.    9th  Ed.     Cr.  8vo.    7s.  6d. 
*^^*  Also  a  Library  Edition.     Demy  Svo.    143. 
THE  AMERICAN  COMMONWEALTH.     2  vols.    Ex.  cr.  Svo.    25s.    Part  I. 

The   National    Government.    Part   II.    The   State  Governments.     Part  III. 

The  Party  System.      Part  IV.  Public  Opinion.     Part  V.    Illustrations  and 

Reflections.     Part  VI.  Social  Institutions. 
*BUOKLEY.-A  HISTORY  OF  ENGLAND  FOR  BEGINNERS.     By  Arabella 

B.  Buckley.    With  Maps  and  Tables.    Gl.  8vo.    3s. 

BURY.— A  HISTORY  OF  THE  LATER  ROMAN  EMPIRE  FROM  ARCADIUS 
TO  IRENE,  A.D.  895-800.  By  John  B.  Bury,  M.A.,  Fellow  of  Trinity  College, 
Dublin.    2  vols.     8vo.    S2s. 

CASSEL.— MANUAL  OF  JEWISH  HISTORY  AND  LITERATURE.  By  Dr.  D. 
Cassel.    Translated  by  Mrs.  Henry  Lucas.    Fcap.  8vo.    2s.  6d. 

ENGLISH  STATESMEN,  TWELVE.    Cr.  8vo.    2s.  6d.  each. 
William  the  Conqueror.    By  Edward  A.  Freeman,  D.C.L.,  LL.D. 
Henry  II.     By  Mrs.  J.  R.  Green. 

Edward  I.    By  P.  York  Powell.  [In  preparation. 

Henry  VII.    By  James  Gairdner. 


48  HISTORY 

Cardinal  Wolsey.    By  Professor  M.  Creighton. 

Elizabeth.     By  B.  S.  Beesly.  [In  preparation. 

Oliver  Cromwell.    By  Frederic  Harrison. 
William  III.     By  H.  U.  Traill. 
Walpole.    By  John  Morley, 

Chatham.    By  John  Morley.  [In  preparation. 

Pitt.    By  John  Morley.  [hi  preparation. 

Peel.    By  J.  R.  Thursfield. 
FISKE.— Works  by  John  Fiske,  formerly  Lecturer  on  Philosophy  at  Harvard 
University. 
THE  CRITICAL  PERIOD   IN  AMERICAN  HISTORY,  1783-1789.      Ex.  cr. 

8vo.     10s.  6d. 
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FREEMAN.— Works  by  Edward  A.  Freeman,  D.C.L.,  Regius  Professor  of  Modem 
History  in  the  University  of  Oxford,  etc. 
*OLD  ENGLISH  HISTORY.    With  Maps.    Ex.  fcap.  8vo.     6s. 
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METHODS  OF  HISTORICAL  STUDY.    8vo.     10s.  6d. 
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^GENERAL  SKETCH  OF  EUROPEAN  HISTORY.    Enlarged,  with  Maps,  etc. 

ISmo.     3s.  6d. 
*PRIMER  OF  EUROPEAN  HISTORY.    ISmo.    Is.    (History  Primers.) 
FRIEDMANN.— ANNE  BOLEYN.    A  Chapter  of  English  History,  1527-1536,    By 

Paul  Friedmann.    2  vols.    Svo.    2Ss. 
FYFFE.— A  SCHOOL  HISTORY  OF  GREECE.    By  C.  A.   Fyffe,   M.A.,   late 
Fellow  of  University  College,  Oxford.    Cr.  Svo.  [In  preparation. 

GIBBINS.— THE  HISTORY  OF    COMMERCE    IN  EUROPE.      By  H.   de  B. 

GiBBiNS,  M.A.     With  Maps.     ISrao.     2s.  6d. 
GREEN.— Works  by  John  Richard  Green,  LL.D.,  late  Honorary  Fellow  of 
Jesus  College,  Oxford. 
*A  SHORT  HISTORY  OF  THE  ENGLISH  PEOPLE.    New  and  Revised  Ed. 
With  Maps,  Genealogical  Tables,  and  Chronological  Annals.    Cr.  Svo.    Ss.  6d. 
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62  DIVINITY 

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The  Patagonia.  By  Henry 
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